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48531 | Find the equation of the line in standard form.
Find the equation of the line that passes through the points [imath](2, 9)[/imath] and [imath](8, 6)[/imath]. Write the equation in standard form. | 657442 | Write the Linear equation
In 1940 there were [imath]245,300[/imath] immigrants admitted to a country. In 2006 there were [imath]1,060,431[/imath]. Write a linear equation expressing the number of immigrants, [imath]y[/imath], in terms of [imath]t[/imath], the number of years after 1900. What is the estimate of immigrants in 2015? |
305810 | Need help in proving a unique invariant borel probability measure is ergodic.
Let [imath]X[/imath] be a topological space and [imath]f : X \rightarrow X[/imath] be a function. Suppose that there exists a unique invariant borel probability measure m. We need show that m is ergodic : I assumed by contradiction that m is not ergodic. Then there is [imath]A[/imath] measurable such that [imath]f^{-1}(A)=A[/imath] and [imath]0 < m(A) < 1[/imath]. How do I go from here?? Thank you for your help. | 305769 | Prove that [imath]m[/imath] is ergodic.
Let [imath]X[/imath] be a topological space, [imath]f\colon X\rightarrow X[/imath] be a function. Suppose that there exists a unique invariant Borel probability measure [imath]m[/imath]. Prove that [imath]m[/imath] is ergodic. Thank you. |
5309 | Topological group: Multiplying two loops is homotopic to linking these paths?
Let G be a topological group and let [imath]s_1[/imath] and [imath]s_2[/imath] be loops in G (both loops are based at the identity e of G). Is it true that the loop [imath]s_1s_2[/imath] (where the multiplication is the one of the group structure of G) is equal, in [imath]\pi_1(G,e)[/imath], to the loop [imath]s_1*s_2[/imath] where this product is given by first going around [imath]s_1[/imath] and then [imath]s_2[/imath] (i.e., do we have [imath][s_1s_2] = [s_1*s_2][/imath])? If yes, what is the proof for this? | 692090 | Fundamental group of a topological group is abelian?
This is related to a problem in Peter May's book concise course in algebraic topology. One is asked to prove that composing loops is equivalent to pointwise multiplication of the loops (problem 3, page 11) and then deduce that [imath]\pi_1[/imath] is abelian. I don't have a clue on how to proceed for the last part (the abelian property) without assuming that the topological group is in fact abelian too. I would appreciate any advice as to how proceed. Thanks |
306513 | Solve the partial differential equation [imath]u_t + uu_x=0[/imath]
Solve the following partial differential equation [imath]u_t + uu_x=0[/imath] with [imath]u=u(x,t)[/imath] and [imath]u(x,0)=x[/imath]. I am having trouble in applying the SIDE CONDITION. The Characteristics are [imath]dx/dt[/imath]=[imath]u[/imath], here u is constant along the characteristics. Along the characteristics [imath]dx/dt=g(x_o)[/imath] Solution is [imath]x=g(X_o)t + d[/imath], where d is an arbitary constant But note at [imath]t=0[/imath], [imath]x=x_0[/imath]...Hence [imath]d=x_0[/imath] The Characteristics is [imath]x=g(x_0)t + x_0[/imath], these are straight lines with variable slope I am not sure on what to do after this point. | 305727 | Solve Burgers' equation
Solve Burgers' equation [imath]u_t + uu_x =0,[/imath] with [imath]u=u(x,t)[/imath] and the side condition [imath]u(x,0)=x[/imath]. I am not sure on how to find the solution [imath]u(x,t)[/imath]. I have learned the method of characteristics. I am neither sure on how to use the side condition in Burgers' equation. |
292326 | log-sine and log-cos integral with fun result
A friend of mine posed the following improper integral: [imath]\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\ln^{2}(\sin(x))\ln^{2}(\cos(x))}{\sin(x)\cos(x)}dx=\frac{1}{2}\zeta(5)-\frac{1}{4}\zeta(2)\zeta(3)[/imath]. This checks numerically. But, does anyone know of a nice place to begin?. I tried the classic series [imath]\displaystyle \ln(\sin(x))=-\sum_{k=1}^{\infty}\frac{\cos(2kx)}{k}-\ln(2)[/imath] and [imath]\displaystyle \ln(\cos(x))=-\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos(2kx)}{k}-\ln(2)[/imath] Then, tried squaring and so forth. It gets rather tedious, but perhaps it'll work given the effort. Then again, the [imath]\sin(x)\cos(x)[/imath] in the denominator must be dealt with. With all of those zetas in the solution, one would think that series may be the way to go. Does anyone know of a nice approach?. Thanks. | 290250 | Show that [imath]\int_{0}^{\pi/2}\frac {\log^2\sin x\log^2\cos x}{\cos x\sin x}\mathrm{d}x=\frac14\left( 2\zeta (5)-\zeta(2)\zeta (3)\right)[/imath]
Show that : [imath] \int_{0}^{\Large\frac\pi2} {\ln^{2}\left(\vphantom{\large A}\cos\left(x\right)\right) \ln^{2}\left(\vphantom{\large A}\sin\left(x\right)\right) \over \cos\left(x\right)\sin\left(x\right)}\,{\rm d}x ={1 \over 4}\, \bigg[2\,\zeta\left(5\right) - \zeta\left(2\right)\zeta\left(3\right) \bigg] [/imath] I can only do non squared one. Anyone has a clue? |
306948 | prove problem if [imath]m^2 | n^2, \text{then } m|n[/imath]
Prove that if [imath]m^2 | n^2, \text{then } m|n[/imath] if [imath]m^2 | n^3, \text{then } m|n[/imath]? prove or disprove. i really no idea of that, someone could help me? | 182988 | If [imath]a^2[/imath] divides [imath]b^2[/imath], then [imath]a[/imath] divides [imath]b[/imath]
Let [imath]a[/imath] and [imath]b[/imath] be positive integers. Prove that: If [imath]a^2[/imath] divides [imath]b^2[/imath], then [imath]a[/imath] divides [imath]b[/imath]. Context: the lecturer wrote this up in my notes without proving it, but I can't seem to figure out why it's true. Would appreciate a solution. |
307385 | How do I differentiate this with respect to x?
I need to differentiate this equation with respect to [imath]x[/imath]: [imath] u(x)=\left[x\left((1-\alpha)+\alpha\left(\frac{\pi(\hat{W}/x-1)}{P}\right)^{\frac{\psi-1}{\psi}}\right)^{\frac{\psi}{\psi-1}}\right]^{\frac{\sigma-1}{\sigma}},[/imath] or [imath]u(x)=\left[x \ h(x) \right] ^{\frac{\sigma-1}{\sigma}},[/imath] where [imath]\alpha\in(0,1)[/imath], [imath]x>0[/imath], [imath]\psi>0[/imath], [imath]\psi\neq 1[/imath] and [imath]\sigma>0[/imath]. [imath]\pi,P,\hat{W}[/imath] are positive constants. I need a solution expressed with the h-function. For example [imath]u'(x) = ?? h(x) ^{??}. [/imath] Hope to hear from someone, thanks in advance | 306680 | Can anyone help me to differentiate this equation?
I need to differentiate this equation with respect to [imath]x[/imath]: [imath] u(x)=\left[x\left((1-\alpha)+\alpha\left(\frac{\pi(\hat{W}/x-1)}{P}\right)^{\frac{\psi-1}{\psi}}\right)^{\frac{\psi}{\psi-1}}\right]^{\frac{\sigma-1}{\sigma}},[/imath] where [imath]\alpha\in(0,1)[/imath], [imath]x>0[/imath], [imath]\psi>0[/imath], [imath]\psi\neq 1[/imath] and [imath]\sigma>0[/imath]. [imath]\pi,P,\hat{W}[/imath] are positive constants. Would be nice if the result can be expressed with a h-function. Let me rewrite the above as [imath]u(x)=\left[x*h(x)\right]^{\frac{\sigma-1}{\sigma}}.[/imath] [imath]u'(x) = ?? h(x) ^{??} [/imath] Hope someone can help, thanks in advance -MB |
307701 | Proof in Group Theory
Show that if [imath]G[/imath] is a finite group with identity [imath]e[/imath] and with an even number of elements, then there is an [imath]a \neq e[/imath] in [imath]G[/imath], such that [imath]a \cdot a = e[/imath]. I read the solutions here http://noether.uoregon.edu/~tingey/fall02/444/hw2.pdf Why do they say [imath]D = \{a, a^\prime\}[/imath]? Isn't [imath]D[/imath] not a group? There is no identity and if they include the identity they get 3 elements, which means [imath]|D| = 3 = [/imath] odd. | 302031 | If G is a finite group with an even number of elements, then binary product of two distinct elements is identity.
Let [imath]G[/imath] be finite group, which has an even number of elements. Show that at least for two (distinct) elements [imath]g,h[/imath] of group [imath]G[/imath] one has [imath]g*g = e[/imath] and [imath]h*h = e[/imath]. I just started learning algebra and I have no ideas how I should solve this. I'm grateful for every explanation. Reference: Fraleigh p. 48 Question 4.29 in A First Course in Abstract Algebra |
307529 | If [imath]\mathrm{char}(K)=p[/imath] is prime, [imath]L/K[/imath] is separable if and only if [imath]K(\alpha) =K(\alpha^p)[/imath] for all [imath]\alpha \in L[/imath]
I am trying to prove that if [imath]L/K[/imath] is an algebraic extension and if [imath]\alpha \in L[/imath], then [imath]\alpha[/imath] is separable over [imath]K[/imath] if [imath]\mathrm{char}(K)=0[/imath]. This is clear because [imath]K[/imath] is perfect which in turn implies that [imath]L/K[/imath] is seperable . Now if [imath]\mathrm{char}(K)=p[/imath] is prime, then the statement is: [imath]\alpha[/imath] is separable if and only if [imath]K(\alpha) =K(\alpha^p)[/imath]. This problem is from a past examination and looks very interesting, but somehow I am not able to connect what's happening with the extension of [imath]K[/imath] with [imath]\alpha[/imath] and [imath]\alpha^p[/imath] . I need some help here. This somehow makes me think that the minimal polynomial of [imath]\alpha[/imath] and [imath]\alpha^p[/imath] have different roots and [imath]\alpha[/imath] is given by some root of minimal polynomial of [imath]\alpha^p[/imath] . Note that I am using the definition of separability as follows : An extension [imath]L/K[/imath] is called separable if for every [imath]\alpha \in L[/imath] the minimal polynomial of [imath]\alpha[/imath] has distinct roots in [imath]L[/imath] . Thanks for helping. | 126242 | Exercise on separable polynomials over fields of prime characteristic
Having learned about separable polynomials today in class, I tried to do the following exercise concerning separable polynomials, namely: Suppose [imath]f[/imath] is the minimal polynomial of [imath]a[/imath] over a field [imath]F[/imath] of prime characteristic. Let [imath]K = F[a][/imath]. Then [imath]f[/imath] is separable iff [imath]F[a^p] = K[/imath]. Now one direction I have proved, that is not separable implies [imath]F[a^p] \subsetneqq F[a][/imath]. For the other direction, I have some trouble at the end( which I will describe). Suppose [imath]f[/imath] is separable and let [imath]g[/imath] be the minimal polynomial of [imath]a[/imath] over [imath]F[a^p][/imath]. We show that [imath]g[/imath] has degree one so that [imath]a \in F[a^p][/imath], proving that [imath]F[a] = F[a^p][/imath]. Suppose we consider [imath]g,f[/imath] as polynomials in [imath]\big(F[a]\big)[x][/imath]. Then [imath]g(a) = f(a) = 0[/imath] in [imath]\big(F[a]\big)[x][/imath]. Since [imath][F[a]:F] > \bigg[F[a]:F[a^p] \bigg],[/imath] this means that [imath]g |f[/imath]. Now write [imath]f(x) = (x-a)u[/imath] where [imath]u[/imath] is some polynomial with coefficients in [imath]F[a][/imath]. Then we observe that since [imath]a[/imath] is algebraic over [imath]F[/imath], [imath]F[a] = F(a)[/imath] that is a field, hence trivially a UFD. Therefore the polynomial ring [imath]\big( F[a] \big)[x][/imath] is a UFD so that [imath]g[/imath] being irreducible is actually prime. Now what I want to do now is to show that if [imath]g|f[/imath], then [imath]g[/imath] must divide [imath](x-a)[/imath] forcing [imath]g = (x-a)[/imath] up to multiplication by a unit. Suppose [imath]g \nmid (x-a)[/imath] so that [imath]g|u[/imath] by [imath]g[/imath] being a prime element in [imath]\big(F[a]\big)[x][/imath]. Since [imath]f[/imath] is separable its derivative is not zero, so that if [imath]g[/imath] divides [imath]f'(x) = (x-a)u' + u[/imath] then I will have my desired contradiction. This is because we will have [imath]g|u[/imath] and [imath]g|u'[/imath] implying that [imath]u[/imath] has a multiple root, contradicting [imath]f[/imath] having no multiple roots. The problem now is how do I know that [imath]g|f'(x)[/imath]? If it is not true here that [imath]g|f'(x)[/imath], can the approach I have done above be salvaged? For reference, the following theorem may be useful: Let [imath]f[/imath] be an irreducible polynomial in [imath]F[x][/imath]. The following are equivalent: (1) [imath]f(x)[/imath] is not separable. (2) [imath]f'(x) = 0[/imath] (3) [imath]\operatorname{Char} F = p >0[/imath] and [imath]f[/imath] is a polynomial in [imath]x^p[/imath] |
203704 | Prove that [imath]a^{p-1} \equiv 1[/imath] [imath]\ [/imath](mod [imath]p[/imath])
Consider the equivalence relation [imath]m\sim n[/imath] defined to be [imath]\frac {m-n}p=z[/imath] (i.e., when [imath]m-n[/imath] is divisible by p) where [imath]m,n,p,z\in \mathbb{Z}[/imath] and [imath]p[/imath] is prime. Now suppose that [imath]a[/imath] is some integer not divisible by [imath]p[/imath]. Prove that [imath][a]^{p-1}=[1][/imath] where [1] denotes the equivalence class of 1. Another common way to write this is [imath]a^{p-1} \equiv 1[/imath] [imath]\ [/imath](mod [imath]p[/imath]). I understand that there is some intent to prove multiplicative inverses here, but I'm a bit confused as to how one would approach it. | 150279 | Proof of Fermat's little theorem using congruence modulo [imath]p[/imath]
I have managed to show that [imath](a + b)^p \equiv a^p + b^p \pmod p[/imath], [imath]a[/imath] and [imath]b[/imath] being any integer and [imath]p[/imath] any prime. How can I prove from this that [imath]a^p \equiv a \pmod p[/imath]? |
307877 | Find the probability mass function of the (discrete) random variable [imath]X = Int(nU) + 1[/imath].
For a non-negative real number [imath]x[/imath], write [imath]Int(x)[/imath] for the largest integer that is less than or equal to [imath]x[/imath]. Let [imath]U[/imath] be a uniform random variable on [imath](0,1)[/imath] and [imath]n \geq 1[/imath] an integer. Find the probability mass function of the (discrete) random variable [imath]X = Int(nU) + 1[/imath]. I don't really know how to start on this question, can someone please give me some hints? | 307481 | Uniform random variable problem
[imath]U[/imath] is a uniform r.v on [imath][0,1][/imath] and [imath]n\geq 1[/imath] is an integer. What is the probability mass function of the (discrete) random variable [imath]X = \lfloor{nU}\rfloor + 1[/imath]? |
308246 | Linear Transformations and Identity Matrix
Suppose that [imath]V[/imath] is a finite dimensional vector space and [imath]T[/imath] is a linear transformation from [imath]V[/imath] to [imath]V[/imath]. Prove that [imath]T[/imath] is a scalar multiple of the identity matrix iff [imath]ST=TS[/imath] for every linear transformation [imath]S[/imath] from [imath]V[/imath] to [imath]V[/imath]. | 284043 | Centre of a matrix ring are [imath] \operatorname{diag}\{ a, a, ..., a \} [/imath] with [imath] a\in Z(R) [/imath]
Show that [imath]Z(M_n(R))[/imath] consist of [imath] \operatorname{diag}\{ a, a, ..., a \} [/imath] with [imath] a\in Z(R) [/imath] |
307099 | Linear Combination of Exponential Random Variables
Let [imath]Y \sim \exp(\delta)[/imath] and [imath]T \sim \exp(\lambda)[/imath], and [imath]Y[/imath] and [imath]T[/imath] are independent. How do I get the density [imath]f(x)[/imath] where [imath]X=Y-cT[/imath], [imath]c>0[/imath]? Thanks. | 115022 | pdf of the difference of two exponentially distributed random variables
Suppose we have [imath]v[/imath] and [imath]u[/imath], both are independent and exponentially distributed random variables with parameters [imath]\mu[/imath] and [imath]\lambda[/imath], respectively. How can we calculate the pdf of [imath]v-u[/imath]? |
308570 | [imath]\sum_{n=1}^{\infty } \left ( 1 - \cos(\frac{1}{n}) \right )[/imath] converges or diverges?
[imath]\sum_{n=1}^{\infty } \left ( 1 - \cos\left(\frac{1}{n}\right) \right )[/imath] I tried all of the tests but I couldnt solve this.Does it converges or diverges? | 155658 | Convergence or divergence of [imath]\sum_{k=1}^{\infty} \left(1-\cos\frac{1}{k}\right)[/imath]
Does [imath]\sum_{k=1}^{\infty} \left(1-\cos\frac{1}{k}\right)[/imath] converge or diverge? |
308930 | Stirling Number Identities
If [imath]\sum_{k=0}^nS(n,k)=B(n)\;,[/imath] the bell number, then does [imath]\sum kS(n,k)=B(n+1)-B(n)\;?[/imath] | 26762 | Formula for [imath]\sum_{k=0}^n S(n,k) k[/imath], where [imath]S(n,k)[/imath] is a Stirling number of the second kind?
I would like to compute [imath]\sum_{k=0}^n S(n,k) k[/imath], where [imath]S(n,k)[/imath] is a Stirling number of the second kind. Any ideas? It is like I am convolving the Stirling numbers of the second kind with the positive integers. Thank you very much! |
290019 | Cardinality of [imath]\mathbb{R}[/imath] and [imath]\mathbb{R}^2[/imath]
I am working on this exercise for an introductory Real Analysis course: Show that |[imath]\mathbb{R}[/imath]| = |[imath]\mathbb{R}^2[/imath]|. I know that [imath]\mathbb{R}[/imath] is uncountable. I also know that two sets [imath]A[/imath] and [imath]B[/imath] have the same cardinality if there is a bijection from [imath]A[/imath] onto [imath]B[/imath]. So if I show that there exists a bijection from [imath]\mathbb{R}[/imath] onto [imath]\mathbb{R}^2[/imath] then I beleive that shows that |[imath]\mathbb{R}[/imath]| = |[imath]\mathbb{R}^2[/imath]|. Let [imath]x_i \in \mathbb{R}[/imath], where each [imath]x_i[/imath] is expressed as an infinite decimal, written as [imath]x_i = x_{i0}.x_{i1}x_{i2}x_{i3}...,[/imath]. Each [imath]x_{i0}[/imath] is an integer, and [imath]x_{ik} \in \left \{ 0,1,2, 3, 4, 5, 6, 7, 8, 9 \right \}[/imath]. Then, let [imath]f(x_i)=(x_{i0}.x_{i1}x_{i3}x_{i5}... ,x_{i0}.x_{i2}x_{i4}x_{i6}...)[/imath] What should I do to show that [imath]f: \mathbb{R} \to \mathbb{R}^2[/imath] is an injective function? Any suggestions or help with the question would be appreciated. | 426503 | Bijection between [imath]\mathbb{R}\times\mathbb{R}[/imath] and [imath]\mathbb{R}[/imath]
It must be posted somewhere, but I can't find it. I've been working on it for a while too without getting anywhere. Does there exist a bijection between [imath]\mathbb{R}\times\mathbb{R}[/imath] and [imath]\mathbb{R}[/imath]? Is it possible to give an explicit bijection? NOTE: This question is not a duplicate of the link suggested. Please see comments for further detail. |
309582 | How to prove [imath]\sum^n_{i=1} \frac{1}{i(i+1)} = \frac{n}{n+1}[/imath]?
How can I prove that [imath]\sum^n_{i=1} \frac{1}{i(i+1)} = \frac{n}{n+1}[/imath]? I noticed that in the sum, the denominator has terms that cancel out, but I'm not sure how to take advantage of that. | 286024 | What is the formula for [imath]\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}[/imath]
How can I find the formula for the following equation? [imath]\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}[/imath] More importantly, how would you approach finding the formula? I have found that every time, the denominator number seems to go up by [imath]n+2[/imath], but that's about as far as I have been able to get: [imath]\frac12 + \frac16 + \frac1{12} + \frac1{20} + \frac1{30}...[/imath] the denominator increases by [imath]4,6,8,10,12,\ldots[/imath] etc. So how should I approach finding the formula? Thanks! |
310273 | Cardinality of the set [imath]\{(A, B) \mid A ⊆ B ⊆ S\}[/imath]
If [imath]S[/imath] is a set with [imath]n[/imath] elements, what is the cardinality of the set [imath]\left\{(A, B) \mid A ⊆ B ⊆ S\right\}[/imath] | 88063 | Cardinality of set containing true order relations from a power set
This question is from the book Introduction to Topology and Modern Analysis by GF Simmons, Problem 3(d) at end of Section 1. I'll paraphrase the question here. Suppose [imath]U[/imath] is a containing [imath]n[/imath] elements. Then, how many subsets are there? How many possible relations of the form [imath]A \subseteq B [/imath] exist where [imath]A,\ B[/imath] is a subset of [imath]U[/imath]? Can you make an informed guess as to how many are true? Solution: First two are simple. [imath]2^n[/imath] and [imath]2^{2n}[/imath]. For the third one, assume [imath]A[/imath] contains [imath]k[/imath] elements. Then, number of different possible [imath]A[/imath] is [imath]\binom{n}{k}[/imath] and the number of possible sets, of which [imath]A[/imath] is a subset, (number of different possible [imath]B[/imath]) is [imath]2^{n-k}[/imath]. Taking sum while varying [imath]k[/imath] from [imath]0[/imath] to [imath]n[/imath] gives us [imath]\sum 2^{n-k} \binom{n}{k} [/imath] which is [imath]3^n[/imath]. Is this correct? If it is? Then why does the book tells us to make an informed guess? Reading the word informed guess made me suspicious that the question is not as easy as it looks. |
310612 | Show that [imath]d_b(x,y)=\frac{d(x,y)}{1+d(x,y)}[/imath] is a metric.
where [imath](X,d)[/imath] is a metric and [imath]x,y \in X[/imath]. I know we need to show: non-negativity: [imath]d(x,y)\geq[/imath] 0 [imath]d(x,y)=0[/imath] if and only if [imath]x=y[/imath] symmetry: [imath]d(x,y)=d(y,x)[/imath] [imath]d(x,z)\leq d(x,y) + d(y,z)[/imath] I think we need to consider the derivative of [imath]f(t)=\frac{t}{1+t}[/imath] however I am not sure what to do next. | 309198 | If [imath]d(x,y)[/imath] is a metric, then [imath]\frac{d(x,y)}{1 + d(x,y)}[/imath] is also a metric
Let [imath](X,d)[/imath] be a metric space and for [imath]x,y \in X[/imath] define [imath]d_b(x,y) = \dfrac{d(x,y)}{1 + d(x,y)}[/imath] a) show that [imath]d_b[/imath] is a metric on [imath]X[/imath] Hint: consider the derivative of [imath]f(t)[/imath] = [imath]\dfrac{t}{1+t}[/imath] b) show that [imath] d[/imath] and [imath] d_b [/imath] are equivalent metrics. c) let [imath](X,d) [/imath] be [imath](\mathbb{R}, |\cdot|)[/imath] Show that there exists no [imath] M>0[/imath] such that [imath] |x-y| [/imath] [imath]\leq[/imath] [imath]Md_b[/imath] [imath](x,y)[/imath] for all [imath] x,y[/imath] [imath]\in[/imath] [imath]\mathbb{R}[/imath] I have calculated [imath]f(t)[/imath] and [imath]f'(t)[/imath] and from this I know [imath]f(t)[/imath] is an increasing function as [imath]f'(t)[/imath] is strictly positive. But I don't know where to go from here or how to do parts b) and c) |
310815 | Which of the following functions are norms?
For [imath]x=(x_1,x_2)[/imath], which of the following functions on [imath]\mathbb{R}^2[/imath] are norms? a.) [imath]A_1(x) = 7\mid x_1\mid + 3\mid x_2\mid[/imath], b.) [imath]A_2(x) = \text{max}\lbrace\mid x_1\mid^2,\mid x_2\mid^2\rbrace[/imath], c.) [imath]A_3(x) = \mid x_1\mid[/imath]. I know there are various rules involved when determining whether a function is a norm, however I am uncertain of how to use utilise them. The rules being scalar multiplication, triangle inequality and zero vectors. | 309111 | Norms Abstract Analysis
I have a question relating to norms and have been giving functions and need to state whether they are norms or not... which of the following are norms on [imath]\mathbb{R}^2[/imath]? Give reasons for your answers. For [imath]x=(x_1,x_2)[/imath] let [imath]G_2 (x) = 7|x_1| + 3|x_2|[/imath] [imath]G_3 (x) = max\left\{|x_1|^2, |x_2|^2\right\}[/imath] I know the criteria for it to be a norm... [imath]||x|| \geq 0[/imath] [imath]||x|| =0[/imath] if [imath]x =0[/imath] [imath]||\lambda x|| = |\lambda|\cdot||x||[/imath] [imath]||x+y||\leq ||x|| + ||y||[/imath] I just cant work out how to use this to prove whether or not they're norms. |
310968 | How many bit strings of length eight contain either three consecutive [imath]0'[/imath]s or four consecutive [imath]1'[/imath]s?
How many bit strings of length eight contain either three consecutive [imath]0'[/imath]s or four consecutive [imath]1'[/imath]s? [imath]2^5 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4[/imath] [imath]2^4 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3[/imath] take the first line plus the second line minus [imath]5![/imath] Would this be the correct answer, if not why? | 178605 | Number of bit strings with 3 consecutive zeros or 4 consecutive 1s
I am trying to count the number of bit-strings of length 8 with 3 consecutive zeros or 4 consecutive ones. I was able to calculate it, but I am overcounting. The correct answer is [imath]147[/imath], I got [imath]148[/imath]. I calculated it as follows: Number of strings with 3 consecutive zeros = [imath]2^5+5\times2^4 = 112[/imath], because the 3 zeros can start at bit number 1, 2, 3, .., 6 Number of strings with 4 consecutive ones = [imath]2^4+4\times2^3 = 48[/imath], I used the same reasoning. Now I am trying to count the number of bit-strings that contain both 3 consecutive zeros and 4 consecutive 1s. I reasoned as follows: the strings can be of the following forms: 0001111x, 000x1111, x0001111..thus there are [imath]2+2+2 = 6[/imath] possibilities for bit-strings where the 3 consecutive zeros come first. Symmetrically there are [imath]6[/imath] bit-strings where the 4 consecutive ones come first. Thus the answer should be = [imath]112+48-12 = 148[/imath]. clearly there's something wrong with my reasoning, if someone could point it out, that would be awesome. Thanks |
311363 | System of roots
Let [imath]\Phi[/imath] an irreducible system of roots, [imath]\Phi^{+} \subset \Phi[/imath] a choose of positive roots. I have to prove that if [imath](\alpha, \beta) \ge 0[/imath] for al [imath]\beta \in \Phi^{+}[/imath] then [imath]\alpha[/imath] is the highest among roots of the same lenght. I have a long way... is there a way that uses only some theorem? | 311687 | Lie algebras and roots systems
Let [imath]\Phi[/imath] an irreducible system of roots, [imath]\Phi^{+} \subset \Phi[/imath] a choose of positive roots. I have to prove that if [imath](\alpha, \beta) \ge 0[/imath] for al [imath]\beta \in \Phi^{+}[/imath] then [imath]\alpha[/imath] is the highest among roots of the same lenght. I have a long way... is there a way that uses only some theorem? My way: Let be [imath]\alpha '\neq \alpha[/imath] the root of maximal lenght [imath]||\alpha||[/imath]. Obviusly [imath]\alpha '\in \Phi_+[/imath], so [imath](\alpha,\alpha')\ge 0[/imath]. \ I want to prove that [imath](\alpha,\alpha')>0[/imath]. \ I show now that [imath]\alpha'=\sum_{\alpha_i\in\Delta}m_i\alpha_i\,\,\,\,\,\,\,\,(m_j>0)[/imath] We can suppose that [imath]\alpha'[/imath] is short. Now we suppose that [imath]\alpha'[/imath] is short. [imath]\Phi[/imath] is irreducible than it exists a simple root [imath]\beta_i[/imath] such that [imath](\alpha',\beta_i)<0[/imath] so [imath]\alpha'+\beta_i[/imath] is a positive root longer than [imath]\alpha'[/imath].\ Morever [imath]||\alpha'+\beta_i||=||\alpha'||[/imath], because [imath](\alpha'+\beta_i,\alpha'+\beta_i)=(\alpha',\alpha')+(\beta_i,\beta_i)+2(\alpha'+\beta_i)=[/imath] [imath] =(\alpha',\alpha')+(\beta_i,\beta_i)\left[1+\frac{2(\alpha',\beta_i)}{(\beta_i,\beta_i)} \right] [/imath] [imath]\leq (\alpha',\alpha')[/imath] Because we have [imath]<\alpha',\beta_i>\leq -1[/imath], for the irreducibility of [imath]\Phi[/imath] we have finish.\ Now if [imath](\alpha,\alpha')=0[/imath], we have [imath]\sum m_j(\alpha,\alpha_j)=0 \,\,\,\,\,\,\,\,\,\,\, (\alpha_j\in\Delta)[/imath] and so [imath](\alpha,\alpha_j)=0 \,\,\,\,\,\,\,\,\,\,\,\,\, (\forall\,\alpha_j\in\Delta)[/imath] and it is no possible.\ We have finally that [imath](\alpha',\alpha)>0[/imath], so [imath]\alpha'-\alpha \in \Phi_+[/imath],because [imath]\alpha maximality[/imath]. However we had seen [imath](\alpha',\alpha)>0[/imath] and [imath](\alpha',\alpha')/(\alpha,\alpha)=1[/imath] we must have [imath]<\alpha',\alpha>=1[/imath]. So we can say that [imath](\alpha',\alpha)=\frac{1}{2}(\alpha,\alpha) [/imath] but for hipotesys we have to have [imath](\alpha,\alpha'-\alpha)\ge 0 [/imath] and so [imath](\alpha,\alpha')-(\alpha,\alpha)=-\frac{1}{2}(\alpha,\alpha)\ge 0 [/imath] and it si no possible. Could you give me a more easy and fast method... maybe using theorems? Is there a method that involves Weyl group? |
311756 | Prove [imath]||a| - |b|| \leq |a - b|[/imath]
I'm trying to prove that [imath]||a| - |b|| \leq |a - b|[/imath]. So far, by using the triangle inequality, I've got: [imath]|a| = |\left(a - b\right) + b| \leq |a - b| + |b|[/imath] Subtracting [imath]|b|[/imath] from both sides yields, [imath]|a| - |b| \leq |a - b|[/imath] The book I'm working from claims you can achieve this proof by considering just two cases: [imath]|a| - |b| \geq 0[/imath] and [imath]|a| - |b| < 0[/imath]. The first case is pretty straightforward: [imath]|a| - |b| \geq 0 \implies ||a| - |b|| = |a| - |b| \leq |a - b|[/imath] But I'm stuck on the case where [imath]|a| - |b| < 0[/imath] Cool, I think I got it (thanks for the hints!). So, [imath]|b| - |a| \leq |b - a| = |a - b|[/imath] And when [imath]|a| - |b| < 0[/imath], [imath]||a| - |b|| = -\left(|a| - |b|\right) = |b| - |a| \leq |a - b|[/imath] | 44504 | How to use triangle inequality to establish Reverse triangle inequality
I need to use [imath]|a+b| \leq |a|+|b|[/imath] to show that [imath]||a|-|b|| \leq |a-b|[/imath] . I have tried to represent [imath]||a|-|b||[/imath] as [imath]||a|+(-|b|)|[/imath] , and then get [imath]||a|+(-|b|)| \leq |a|+|-|b||[/imath] , but that isn't leading me anywhere given [imath]|a-b| \leq |a|+|b|[/imath]. |
311794 | Find total number of mappings
Let [imath]A[/imath] be a set of [imath]n[/imath] elements and [imath]B[/imath] be a set of [imath]m[/imath] elements. Show that the total number of mappings from [imath]A[/imath] to [imath]B[/imath] is [imath]m^n[/imath]. | 223240 | How many distinct functions can be defined from set A to B?
In my discrete mathematics class our notes say that between set [imath]A[/imath] (having [imath]6[/imath] elements) and set [imath]B[/imath] (having [imath]8[/imath] elements), there are [imath]8^6[/imath] distinct functions that can be formed, in other words: [imath]|b|^{|a|}[/imath] distinct functions. But no explanation is offered and I can't seem to figure out why this is true. Can anyone elaborate? |
297443 | Prove the following equality: [imath]\sum_{k=0}^n\binom {n-k }{k} = F_n[/imath]
I need to prove that there is the following equality: [imath] \sum\limits_{k=0}^n {n-k \choose k} = F_{n} [/imath] where [imath]F_{n}[/imath] is a n-th Fibonacci number. The problem seems easy but I can't find the way to prove it. Could you give me any hints? I'm looking for a combinatorial proof. | 292940 | finding the combinatorial sum
How to find the sum of combinatorial summation of the following series, where [imath]C(n, k)[/imath] denotes the number of combinations of [imath]n[/imath] given [imath]k[/imath] are the same? [imath]\sum_{k=0}^{n/2} C(n-k, k)[/imath] Need help on this. |
312174 | What is the automorphism group of [imath]\mathbb{R}[/imath] under addition?
I do not know much about infinite groups. [imath]\mathbb{R}[/imath] is especially different from others I have worked before - it does not seem to have any generator like [imath]\mathbb{Z}[/imath] does or we could say that its every non-trivial element generates a subgroup isomorphic to [imath]\mathbb{Z}[/imath]. I attempted to find the automorphism group of [imath]\mathbb{R}[/imath]. There are only three kinds of automorphism operations we can perform on [imath]\mathbb{R}[/imath]: identity: [imath]\psi_1: x \mapsto x[/imath], [imath]\psi_1 = \mathrm{id}[/imath] reflection: [imath]\psi_2:x \mapsto -x[/imath], [imath]\psi_2 \circ \psi_2 = \mathrm{id}[/imath] translation: [imath]\phi_r:x \mapsto x+r[/imath], [imath]r \in (-\infty, \infty) = \mathbb{R}[/imath] Therefore the [imath]\mathrm{Aut}(\mathbb{R}) = \mathbb{R} * \mathbb{Z}_2[/imath]. Is my reasoning correct? | 115486 | What is [imath]\operatorname{Aut}(\mathbb{R},+)[/imath]?
I was solving some exercises about automorphisms. I was able to show that [imath]\operatorname{Aut}(\mathbb{Q},+)[/imath] is isomorphic to [imath]\mathbb{Q}^{\times}[/imath]. The isomorphism is given by [imath]\Psi(f)=f(1)[/imath], but when I try to do the same thing with [imath]\operatorname{Aut}(\mathbb{R},+)[/imath] I got stuck. My question is: What is [imath]\operatorname{Aut}(\mathbb{R},+)[/imath]? I would appreciate your help. |
312365 | If [imath]P(x)[/imath] is a polynomial then [imath]\lim_{x \to \infty }P(x) = 0 \iff P(x)=0[/imath],
How to show that if [imath]P(x)[/imath] is a polynomial then [imath]\lim_{x \to \infty }P(x) = 0 \iff P(x)=0[/imath], This question was previously posted answered and then deleted. | 23555 | Do all polynomials with order [imath]> 1[/imath] go to [imath]\pm[/imath] infinity?
Background As background, I have found that taylor expansion provides poor estimates of a function at extreme parameter values. Indeed, the approximation at extreme values can get worse (more rapid exponential increase) as the order of the taylor series increases. This seems intuitive, but I don't know that it is a rule... or if there is a proof. Questions Do all polynomials with order [imath]> 1[/imath] go to [imath]\pm[/imath] infinity? Is there a good reference where I can find answers to a qustion such as this? |
125995 | Projective dimension of the residue field of a noetherian local ring.
Let [imath]R[/imath] be a commutative Noetherian local ring with maximal ideal [imath]\mathfrak m[/imath]. Is it true that the projective dimension of [imath]R/\mathfrak m[/imath] is finite knowing that its injective dimension is finite? If yes, why?!? I would need this to prove something else but I'm not sure I can use it. Can you help me please? Thanks. | 110488 | Finite injective dimension of the residue field implies that the ring is regular
Let [imath](R,\mathfrak m,k)[/imath] be a noetherian local ring. If [imath]\operatorname{inj dim}_R k[/imath] is finite, then [imath]R[/imath] is regular. This is exercise 3.1.26 from Bruns and Herzog, Cohen-Macaulay Rings. I don't see how I can use the results from this chapter to solve it. I think we must use the Ext long exact sequence, but I don't see how. |
312874 | Linear Independence of an infinite set 2.
Let [imath]e_n=\sin nx[/imath] ([imath]x\in [-\pi,\pi])[/imath]and let [imath]A=\{e_i|i\in \mathbb{N}\}[/imath]. Prove that A is a linearly independent set. Some hours back I had posted this question. In the meantime I was trying myself to come up with a proof. Now I have some idea which I am be writing down as an answer. I will ask all of you to evaluate whether this is the correct approach or not? | 312646 | Linear Independence of an infinite set .
Let [imath]e_n=\sin nx[/imath] ([imath]x\in [-\pi,\pi])[/imath]and let [imath]A=\{e_i|i\in \mathbb{N}\}[/imath]. Prove that A is a linearly independent set. |
312929 | find the coordinates from a distance matrix
I want to determine whether there exists [imath]5[/imath] points in [imath]\mathbb R^4[/imath] such that the following matrix is the distance matrix. [imath] \begin{pmatrix} 0& \sqrt5 & \sqrt5 & \sqrt5 & \sqrt5 \\ \sqrt5 & 0& 2\sqrt5 & 2\sqrt2 & 2\\ \sqrt5 &2\sqrt5 & 0 & 2 & 2\sqrt2 \\ \sqrt5 & 2\sqrt2 & 2 & 0 &2 \sqrt5 &\\ \sqrt5 & 2 & 2\sqrt2 & 2 \sqrt5 &0\\ \end{pmatrix}.[/imath] I tried to assume that [imath]x_0=(0,0,0,0)[/imath], [imath]x_1=(x_{11},0,0,0)[/imath], [imath]x_2=(x_{21}, x_{22}, 0,0)[/imath], etc.. but I got stuck because I got [imath]x_{22}=0[/imath] | 311875 | Find out whether a matrix is a distance matrix or not
I have a [imath]5 \times 5[/imath] symmetric matrix [imath]A[/imath] with zeroes on the diagonal and I am supposed to find whether there exist 5 points in [imath]\mathbb R^4[/imath] such that [imath]A[/imath] is the distance matrix. How can I solve this formally? I am guessing I have to use a Gram matrix but I don't know what to do after that... Here is the matrix: [imath]A= \begin{pmatrix} 0 &1 &2 &2 &2\sqrt 2\\ 1& 0 &\sqrt5 &\sqrt5 &3 \\ 2 &\sqrt5 &0 &2\sqrt2 &2\\ 2 &\sqrt5 & 2\sqrt2 &0 &2\sqrt3\\ 2\sqrt2 & 3 &2 &2\sqrt3 &0 \end{pmatrix}.[/imath] Edit. I got stuck when I tried to use the technique proposed by user1551 on this matrix [imath] \begin{pmatrix} 0& \sqrt5 & \sqrt5 & \sqrt5 & \sqrt5 \\ \sqrt5 & 0& 2\sqrt5 & 2\sqrt2 & 2\\ \sqrt5 &2\sqrt5 & 0 & 2 & 2\sqrt2 \\ \sqrt5 & 2\sqrt2 & 2 & 0 &2 \sqrt5 &\\ \sqrt5 & 2 & 2\sqrt2 & 2 \sqrt5 &0\\ \end{pmatrix}.[/imath] The problem arises when I try to compute [imath]x_3[/imath]. I obtained the following system: [imath]-2\begin{pmatrix} \sqrt5 & 0\\ -\sqrt5 &0 \\ \end{pmatrix} \begin{pmatrix} x_{31}\\ x_{32}\\ \end{pmatrix} =\begin{pmatrix} 10\\ -1\\ \end{pmatrix}.[/imath] |
312833 | A question on power set of A
I want to prove that [imath]\{f|f:A\rightarrow\{0,1\}\} \approx P(A)[/imath] where P(A) is the power set of A. I think the idea is to show that there is a monotonic function from the set A to its power set. But I cannot go any further. | 84180 | Finding a correspondence between [imath]\{0,1\}^A[/imath] and [imath]\mathcal P(A)[/imath]
I got this question in homework: Let [imath]\{0,1\}^A[/imath] the set of all functions from A (not necessarily a finite set) to [imath]\{0,1\}[/imath]. Find a correspondence (function) between [imath]\{0,1\}^A[/imath] and [imath]\mathcal P(A)[/imath] (The power set of [imath]A[/imath]). Prove that this correspondence is one-to-one and onto. I don't know where to start, so I need a hint. What does it mean to find a correspondence? I'm not really supposed to define a function, right? I guess once I have the correspondence defined somehow, the proof will be easier. Any ideas? Thanks! |
313376 | Linear algebra problem - how to prove this proposition
Let [imath]L:\mathbb{R}^{n}\to\mathbb{R}^{m}[/imath] be a linear mapping. Assume that [imath]\{\vec{v}_{1},\ldots,\vec{v}_{n}\}[/imath] is a basis for [imath]\mathbb{R}^{n}[/imath] such that [imath]\{\vec{v}_{1},\ldots,\vec{v}_{k}\}[/imath] is a basis for [imath]\ker(L)[/imath] . Prove that [imath]\{L(\vec{v_{k+1}}),\ldots,L(\vec{v_{n}})\}[/imath] is linearly independent. I'm trying to prove by contradiction, but I'm stuck on needing to prove [imath]L[/imath] is one-to-one over [imath]\text{Span}\{\vec{v}_{k+1},\dots,\vec{v}_{n}\}[/imath] . Also, the proof is getting really long, so is there a better way to prove the proposition? | 311369 | L : [imath]\mathbb{R}^n \rightarrow \mathbb{R}^m[/imath] is a linear mapping, linear independence of [imath]L[/imath] mapped onto a set of vectors.
Here is my question: Let [imath]L:\mathbb{R}^n\to \mathbb{R}^m[/imath] be a linear mapping. Assume that [imath]\{\vec{v}_1,\ldots,\vec{v}_n\}[/imath] is a basis for [imath]\mathbb{R}^n[/imath] such that [imath]\{\vec{v}_1,\ldots,\vec{v}_k\}[/imath] is a basis for Ker[imath](L)[/imath]. Is [imath]\{L(\vec{v}_{k+1}),\ldots,L(\vec{v}_n)\}[/imath] is linearly independent? I'm trying to write a proof but am having a bit of trouble .. |
313375 | Monotonic function non-continuous in each rational
How can I prove that exists a monotonic non-decreasing function [imath]f: [0,1] \rightarrow \mathbb R[/imath] that isn't continuous in every rational of its domain? | 172753 | Is there a monotonic function discontinuous over some dense set?
Can we construct a monotonic function [imath]f : \mathbb{R} \to \mathbb{R}[/imath] such that there is a dense set in some interval [imath](a,b)[/imath] for which [imath]f[/imath] is discontinuous at all points in the dense set? What about a strictly monotonic function? My intuition tells me that such a function is impossible. Here is a rough sketch of an attempt at proving that such a function does not exist: we could suppose a function satisfies these conditions. Take an [imath]\epsilon > 0[/imath] and two points [imath]x,y[/imath] in this dense set such that [imath]x<y[/imath]. Then, [imath]f(x)<f(y)[/imath] because if they are equal, then the function is constant at all points in between, and there is another element of [imath]X[/imath] between [imath]x[/imath] and [imath]y[/imath], which would be a contradiction. Take [imath]f(y)-f(x)[/imath]. By the Archimedean property of the reals, [imath]f(y)-f(x)<n\epsilon[/imath] for some [imath]n[/imath]. However, after this point, I am stuck. Could we somehow partition [imath](x,y)[/imath] into [imath]n[/imath] subintervals and conclude that there must be some point on the dense set that is continuous? |
287434 | Is the set of all functions from [imath]\mathbb{N}[/imath] to [imath]\{0,1\}[/imath] countable or uncountable?
Is the set of all functions from [imath]\mathbb{N}[/imath] to [imath]\lbrace 0,1\rbrace[/imath] countable or uncountable ? I have no idea on how to start. Any hint or guide will be much appreciated. | 314682 | Is the sets of all maps from [imath]\mathbb{N}[/imath] to [imath]\mathbb{N}[/imath] countable?
Is the sets of all maps from [imath]\mathbb{N}[/imath] to [imath]\mathbb{N}[/imath] countable? Attempts: I think it is uncountable. Consider [imath]P(\mathbb{N})[/imath] where [imath]P(A)[/imath] denote as the power sets of A. We know that [imath]P(\mathbb{N})[/imath] is uncountable. Moreover, by the axiom of choice, for any non empty set, we know that there exist a function [imath]f: P(\mathbb{N})\to \mathbb{N}[/imath] . Note that [imath]P(\mathbb{N})[/imath] have uncountable many elements and since each [imath]f[/imath] belongs to the set of all mapping from [imath]\mathbb{N}[/imath] to [imath]\mathbb{N}[/imath] so there exist infinite uncountably elements and hence this set are uncountable. Is my attempt work?? |
313459 | Why is Hausdorff metric defined this way?
From Wikipedia The definition of the Hausdorff distance can be derived by a series of natural extensions of the distance function [imath]d(x, y)[/imath] in the underlying metric space [imath]M[/imath], as follows:[4] Define a distance function between any point [imath]x[/imath] of [imath]M[/imath] and any non-empty set [imath]Y[/imath] of [imath]M[/imath] by: [imath] d(x,Y)=\inf \{ d(x,y) | y \in Y \}\ . [/imath] Define a distance function between any two non-empty sets [imath]X[/imath] and [imath]Y[/imath] of [imath]M[/imath] by: [imath] d(X,Y)=\sup \{ d(x,Y) | x \in X \}\ . [/imath] If [imath]X[/imath] and [imath]Y[/imath] are compact then [imath]d(X,Y)[/imath] will be finite; [imath]d(X,X)=0[/imath]; and [imath]d[/imath] inherits the triangle inequality property from the distance function in [imath]M[/imath]. As it stands, [imath]d(X,Y)[/imath] is not a metric because [imath]d(X,Y)[/imath] is not always symmetric, and [imath]d(X,Y) = 0[/imath] does not imply that [imath]X = Y[/imath] (It does imply that [imath]X \subseteq Y[/imath]). However, we can create a metric by defining the Hausdorff distance to be: [imath] d_{\mathrm H}(X,Y) = \max\{d(X,Y),d(Y,X) \} \, . [/imath] I was wondering why in the three steps, [imath]\inf[/imath] and [imath]\sup[/imath] are as they are? What if some or all of they are flipped between [imath]\inf[/imath] and [imath]\sup[/imath]? For example, the first step defines the distance between a point and a set by [imath]\inf[/imath], but the second step defines the distance between two sets by [imath]\sup[/imath] and the third step uses [imath]\max[/imath] again. Why are they not all using [imath]\inf[/imath] or [imath]\min[/imath] (or [imath]\sup[/imath] or [imath]\max[/imath], respectively), which I think would make the three steps more agreeable, since in the first step, a point can be seen as a singleton set. Thanks and regards! | 183601 | On the definition of the Hausdorff distance
[imath]\newcommand{\dist}{\mathrm{dist}\,}[/imath] Let [imath]M[/imath] be a metric space and [imath]\emptyset\neq A,B\subset M[/imath] bounded closed subsets. The Hausdorff distance is defined as [imath]h(A,B)=\max\{\dist(A,B),\dist(B,A)\},[/imath] where [imath]\dist(A,B)=\sup_{x\in A}\inf_{y\in B}d(x,y).[/imath] Why do we define [imath]\dist(A,B)[/imath] in this way? Is't it possible to replace the supremum by the infimum in the definition of [imath]\dist\![/imath], that is, define [imath]\dist_{\mathrm{new}}(A,B)=\inf_{x\in A}\inf_{y\in B}d(x,y).[/imath] What is the impact of this 'new' definition on the 'Hausdorff distance' given by [imath]h_{\mathrm{new}}(A,B)=\max\{\dist_{\mathrm{new}}(A,B),\dist_{\mathrm{new}}(B,A)\}?[/imath] |
313735 | If Y is compact, then the projection map of [imath]X \times Y[/imath] is a closed map.
This is a question from munkres section 26 problem 7. Show if Y is compact, then the projection [imath]\pi_1:X \times Y \rightarrow X[/imath] is a closed map. My question is why this is not trivial. Essentially we want that if [imath]C_X \times C_Y[/imath] is a closed subset of [imath]X \times Y[/imath] then, [imath]\pi_1(C_X \times C_Y)=C_X[/imath] is closed. But if [imath]C_X \times C_Y[/imath] is closed, then is it not the case that both [imath]C_X [/imath] and [imath] C_Y[/imath] must be closed? Therefore [imath]C_X[/imath] must be closed. Thus [imath]\pi_1[/imath] is a closed map. Why do we even need [imath]Y[/imath] compact? | 22697 | Projection map being a closed map
Let [imath]\pi: X \times Y \to X[/imath] be the projection map where [imath]Y[/imath] is compact. Prove that [imath]\pi[/imath] is a closed map. First I would like to see a proof of this claim. I want to know that here why compactness is necessary or do we have any other weaker condition other than compactness for the same result to hold. |
313719 | Geometry with vectors
Let [imath]V = 25i + 15j - 30k[/imath]. What angles does this vector make with the [imath]x[/imath], [imath]y[/imath], and [imath]z[/imath] axes? Can someone go all the way through and show me how this is done? Thanks! | 313642 | Vector geometry
Let [imath]V = 25i + 15j - 30k[/imath] . What angles does this vector make with the [imath]x, y[/imath], and [imath]z[/imath] axes? I'm a bit confused on how I should approach this problem. I am trying to find, [imath]\theta_x[/imath], [imath]\theta_y[/imath] and [imath]\theta_z[/imath]. Can anyone lend a hand? |
314079 | Explain why the residue theorem works
I'm studying for a midterm and my teacher warned this would be a good question to understand for the test. The problem is, I do not know how to go about explaining it. Suppose [imath]g(x)[/imath] has a pole of order [imath]2[/imath] at the point z. Explain why: [imath]\textrm{Res}(g;z)= \lim_{x\to z} \frac{d}{dx}((x-z)^2 g(x))[/imath] | 314031 | Explain why the residue is equal to the limit?
I'm studying for a midterm and my teacher warned this would be a good question to understand for the test. The problem is, I do not know how to go about explaining it. Suppose g(x) has a pole of order 2 at the point z. Explain why: [imath]\textrm{Res}(g;z)= \lim_{x\to z} \frac{d}{dx}((x-z)^2 g(x))[/imath] |
314145 | If there are continuous surjections [imath]f: X \to Y[/imath] and [imath]g: Y \to X[/imath], are [imath]X, Y[/imath] homeomorphic?
If there are continuous surjections [imath]f: X \to Y[/imath] and [imath]g: Y \to X[/imath] are [imath]X, Y[/imath] homeomorphic? This is not homework. Just to trying to review some topology. | 169468 | Can two topological spaces surject onto each other but not be homeomorphic?
Let [imath]X[/imath] and [imath]Y[/imath] be topological spaces and [imath]f:X\rightarrow Y[/imath] and [imath]g:Y\rightarrow X[/imath] be surjective continuous maps. Is it necessarily true that [imath]X[/imath] and [imath]Y[/imath] are homeomorphic? I feel like the answer to this question is no, but I haven't been able to come up with any counter example, so I decided to ask here. |
314490 | How to show that the only connected subsets of [imath]\mathbb{Q}[/imath] are the one-point sets?
I need to show that the only connected subsets of [imath]\mathbb{Q}[/imath] are the one-point sets. [imath]\mathbb{Q}[/imath] is given the relative topology of [imath]\mathbb{R}[/imath] | 120842 | What are all of the connected subsets of [imath]\mathbb{Q}[/imath]?
The answer are the singletons of [imath]\mathbb{Q}[/imath]. I can show that the open intervals of [imath]\mathbb{Q}[/imath] are disconnected by choosing some irrational in the open set and using it to form a separation. But strangely enough, I am having a hard time seeing why the subset [imath]\{p,q \}[/imath] of [imath]\mathbb{Q}[/imath] is disconnected. If say [imath]p < q[/imath] and we choose some irrational [imath]p < x <q[/imath], the separation of [imath]\{p,q \}[/imath] would, I think, be given by [imath] \{p\}[/imath] and [imath]\{q\}[/imath]. But these aren't open in [imath]\mathbb{Q}[/imath] ...? |
314766 | What functions satisfy [imath]\sum _{n=1} ^{\infty} x_n < \infty \ \Rightarrow \sum _{n=1} ^{\infty} f(x_n) < \infty [/imath]
Could you help me with this problem? What functions [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] satisfy the following implication? [imath]\sum _{n=1} ^{\infty} x_n < \infty \ \Rightarrow \sum _{n=1} ^{\infty} f(x_n) < \infty [/imath] The necessary condition for convergence of series [imath]\sum _{n=1} ^{\infty} a_n [/imath] is that [imath]\lim a_n =0[/imath]. So I think that [imath]f[/imath] should be decreasing. Series is convergent [imath]\iff[/imath] [imath]\forall \epsilon>0 \ \ \exists N\in \mathbb{N} \ : \ \forall q \ge p \ge k \ : \ |\sum _{n=p} ^q a_n|< \epsilon[/imath]. But I don't know how to use it. Could you help me? Thank you. | 116964 | The set of functions which map convergent series to convergent series
Suppose [imath]f[/imath] is some real function with the above property, i.e. if [imath]\sum\limits_{n = 0}^\infty {x_n}[/imath] converges, then [imath]\sum\limits_{n = 0}^\infty {f(x_n)}[/imath] also converges. My question is: can anything interesting be said regarding the behavior of such a function close to [imath]0[/imath], other than the fact that [imath]f(0)=0[/imath]? |
315080 | Convert [imath]2^{25}[/imath] to its equivalent in base 10 (i.e. [imath]10^?[/imath])
Is there a straightforward way to convert from one number to the power of something to its equivalent as another number to the power of something? For example, how do you convert [imath]2^{25}[/imath] into [imath]10^{something}[/imath]? | 197365 | How to convert this number?
How to convert this number: [imath](-1)\times1.625\times2^{35}[/imath] To this notation: [imath]−5.5834\times10^{10}[/imath] What are the rules for the conversion? This is taken from a Youtube video about single-precision floating point IEEE 754 representation standard. |
315153 | Find [imath]\lim_{n \to \infty} \int_{0}^n\frac{1}{n+n^2 \sin \left( \frac{x}{n^2} \right)} \mbox{d}x [/imath]
Find [imath]\lim_{n \to \infty} \int_{0}^n\frac{1}{n+n^2 \sin \left( \dfrac{x}{n^2} \right)} \mbox{d}x [/imath] I've defined [imath]f_n(x) = \begin{cases} \dfrac{1}{n+n^2 \sin \left( \frac{x}{n^2} \right)}, & \text{if} \ \ x \in [0,n] \\ 0, & \text{if} \ x >n\end{cases}[/imath] Of course [imath]f_n \to 0[/imath]. I have to find [imath]\displaystyle\lim_{n \to \infty} \int_{0}^\infty f_n(x) \mbox{d}x [/imath], I'm trying to use Lebesgue theorem but i can't find function [imath]g[/imath] such that [imath]|f_n(x)| \le g(x)[/imath] and [imath]\displaystyle\int _0 ^\infty g(x) \mbox{d}x < \infty[/imath]. | 313512 | Limit of some integral
My question is how to find: [imath]\displaystyle \lim_{n \rightarrow \infty} \int\limits_0^n \frac {1}{n+n^2\sin(xn^{-2})} dx [/imath]? I've tried with a dominated convergence theorem, but it didn't work. Now, I how absolutely no idea what I can do to solve it. Please, help me. |
315302 | Complete metric space and convergence
I hope someone can help me with the following exercise. I have no clue how to solve this, and I'm very week in topology. If we let (X,d) be a complete metric space and U[imath]\subseteq[/imath] X, U[imath]\neq[/imath] X, it is open subset. Define a function: [imath]\rho: U \times U \rightarrow [0,\infty)[/imath] as: [imath]\rho(x,y):=d(x,y)+ |\frac{1}{d(x,V}+\frac{1}{d(y,V)}|[/imath], where V=X\U where d(x,X\U) is the usual distance between point x and subset X\U. d(x,X\U)=inf{d(x,z)|z [imath]\in[/imath] X\U}. 1) Show that function [imath]\rho[/imath] satisfies the axioms of a metric. 2) Let [imath](x_n)[/imath] be a sequence in U and w[imath]\in[/imath] U. Show that the sequence [imath](x_n)[/imath] converges to w in metric d if and only if it converges to w in metric [imath]\rho[/imath]. Thus, the two metrics d and [imath]\rho[/imath] give rise to the same topology on U. 3) Let ([imath]x_n[/imath]) be a sequence in U, which is Cauchy with respect to metric [imath]\rho[/imath]. Show that [imath](x_n)[/imath] is also Cauchy with respect to metric d, and thus it converges to some point y [imath]\in[/imath] X. Show that y [imath]\in[/imath]U, since otherwise the sequence [imath](x_n)[/imath] would be unbounded with respect to metric [imath]\rho[/imath]. Conclude that (U,[imath]\rho[/imath]) is a complete metric space. If someone could give me some explaining with a lot of details, it would be helpful. \Thanks. | 314195 | Topology - axioms of metric space - convergency - Cauchy
Let [imath](X,d)[/imath] be a complete metric space and [imath]U \subseteq X[/imath], [imath]U \neq X[/imath], its open subset. Define a function [imath]\rho\colon U \times U \rightarrow [0, \infty)[/imath] as: [imath]\rho(x,y):=d(x,y)+\left|\frac{1}{d(x,X\setminus U)} - \frac{1}{d(y,X\setminus U)}\right|,[/imath] where [imath]d(x,X\setminus U)[/imath] is the usual distance between point [imath]x[/imath] and subset [imath]X\setminus U[/imath]: i) Show that function [imath]\rho[/imath] satisfies the axioms of a metric. ii) Let [imath](x_n)[/imath] be a sequence in [imath]U[/imath] and [imath]w \in U[/imath]. Show that the sequence [imath](x_n)[/imath] converges to [imath]w[/imath] in metric [imath]d[/imath] if and only if it converges to [imath]w[/imath] in metric [imath]\rho[/imath]. Thus, the two metrics [imath]d[/imath] and [imath]\rho[/imath] give rise to the same topology on [imath]U[/imath]. iii) Let [imath](x_n)[/imath] be a sequence in [imath]U[/imath], which is Cauchy with respect to metric [imath]\rho[/imath]. Show that [imath](x_n)[/imath] is also Cauchy with respect to metric [imath]d[/imath], and thus it converges to some point [imath]y \in X[/imath]. Show that [imath]y \in U[/imath], since otherwise the sequence [imath](x_n)[/imath] would be unbounded with respect to metric [imath]\rho[/imath]. Conclude that [imath](U,\rho)[/imath] is a complete metric space. I need help for i), ii) and iii). I don't know how to solve them. |
63219 | Nonabelian semidirect products of order [imath]pq[/imath]?
I just constructed the semidirect product in Lang, and I'm trying to tie some facts together. From Ash's Algebra, I know that if [imath]p\lt q[/imath] are distinct primes, if [imath]q\not\equiv 1\pmod{p}[/imath], then any group [imath]G[/imath] of order [imath]pq[/imath] is abelian. Is the converse true, that for any primes [imath]p\lt q[/imath], if [imath]q\equiv 1\pmod{p}[/imath] then there exists a nonabelian group of order [imath]pq[/imath]? One example I found online is that [imath]\mathbb{Z}_3\ltimes \mathbb{Z}_7[/imath] is nonabelian, and here [imath]7\equiv 1\pmod{3}[/imath]. I was considering then semidirect products [imath]\mathbb{Z}_p\ltimes\mathbb{Z}_q[/imath] where [imath]q\equiv 1\pmod{p}[/imath] and some homomorphism [imath]\phi\colon \mathbb{Z}_p\to\operatorname{Aut}(\mathbb{Z}_q)[/imath] I calculate that [imath] (1,0)(0,1)=(1+0,\phi_0(0)+1)=(1,1) [/imath] and [imath] (0,1)(1,0)=(0+1,\phi_{-1}(1)+0)=(1,\phi_{p-1}(1)). [/imath] Is it true somehow that [imath]\phi_{p-1}(1)\neq 1[/imath] in each case to show the group is nonabelian? I guess if it did this would imply [imath]\phi_{p-1}[/imath] is the trivial automorphism, so maybe there's something there? If not, is there a way to show [imath]\mathbb{Z}_p\ltimes\mathbb{Z}_q[/imath] is nonabelian in these cases in general? Thanks. | 1712037 | Let [imath]p be distinct prime numbers and G be a group with |G|=pq[/imath]
Let [imath]p < q[/imath] be distinct prime numbers and [imath]G[/imath] be a group with [imath]|G| = pq[/imath]. Give an example of [imath]p[/imath], [imath]q[/imath] and [imath]G[/imath] such that [imath]G[/imath] is not isomorphic to [imath]\mathbb{Z}_{pq}[/imath]. Now suppose that [imath]p = 5[/imath] and [imath]q = 7[/imath]. Show that [imath]G[/imath] is isomorphic to [imath]\mathbb{Z}_{35}[/imath]. I know we are learning about Sylow's Theorems and group actions, but those are very confusing to me, so I'm not sure how to implement them really. |
316159 | Prove that for every natural number [imath]n[/imath] we have [imath]\gcd(n! + 1, (n + 1)! + 1) = 1[/imath]
I know that we are gonna need to use one of the identities that the [imath]\gcd[/imath] is equal to but I can't remember one that would be useful for this problem. Any help? | 6382 | How to show that [imath]\gcd(n! + 1, (n + 1)! + 1) \mid n[/imath]?
Let [imath]n[/imath] be a positive integer, [imath]n![/imath] denotes the factorial of [imath]n[/imath]. Let [imath]d = \gcd(n! + 1, (n + 1)! + 1)[/imath]. Show that [imath]d[/imath] divides [imath]n[/imath]. (Hint: notice that [imath](n+1)(n!+1) = (n+1)!+n+1[/imath]) |
316305 | Two finite fields are isomorphic.
Let [imath]F = \Bbb{Z}_2[/imath]. Given the irreducible polynomials [imath]f(x)= x^3 + x + 1[/imath], and [imath]g(y) = y^3 + y^2 + 1[/imath], form the fields [imath]K = F[x]/(f(x))[/imath] and [imath]E = F[y] / (g(y))[/imath]. These are fields of order 8 (given), so they must be isomorphic. Is the map [imath][x] \mapsto [y + 1][/imath] an isomorphism? It's clearly onto, and and it's one-one since [imath]F[/imath] and [imath]E[/imath] both have 8 elements. | 91404 | Constructing an explicit isomorphism between finite extensions of finite fields
Suppose [imath]K[/imath] is a finite field, [imath]K = \mathbb F_{p^s}[/imath]. If we take an irreducible polynomial [imath]f[/imath] of degree [imath]d[/imath] over [imath]K[/imath], then the splitting field [imath]L[/imath] of [imath]f[/imath] is [imath]K(\alpha)[/imath] where [imath]f[/imath] is the minimal polynomial of [imath]\alpha[/imath]. But then [imath]L = \mathbb F_{p^{sd}} [/imath]. Since [imath]\mathbb F_{p^{sd}}[/imath] is unique, we see that this is the splitting field of every irreducible polynomial of degree [imath]d[/imath] over [imath]K[/imath]. Take [imath]K = \mathbb F_2[/imath] and let [imath]P(X) = X^3 + X + 1[/imath], [imath]Q(X) = X^3 + X^2 + 1[/imath]. Let [imath]L[/imath] be the splitting field of [imath]P[/imath] and [imath]L'[/imath] be the splitting field of [imath]Q[/imath]. The above tells us that [imath]L[/imath] and [imath]L'[/imath] are isomorphic. I would like to construct an explicit isomorphism between [imath]L[/imath] and [imath]L'[/imath]. I know that [imath]L \cong \mathbb F_2[X] /(X^3 +X + 1)[/imath] and [imath]L' \cong \mathbb F_2[X] / (X^3 + X^2 + 1)[/imath]. Intuitively, I want to find an isomorphism [imath]\phi : \mathbb F_2[X] \to \mathbb F_2[X][/imath] such that [imath]\phi((X^3 + X + 1)) = (X^3 + X^2 + 1)[/imath]. A little playing around gives me [imath]\phi(X) = X+1[/imath]. It now feels like I'm falling at the last hurdle: how do I finish the construction of an isomorphism between [imath]L[/imath] and [imath]L'[/imath]? I don't think [imath]\phi[/imath] makes sense as a map from [imath]L[/imath] to [imath]L'[/imath], yet it seems the map I want. |
316530 | Comparing the two cardinals [imath]\aleph_0^{\aleph_0}[/imath] and [imath]2^{\aleph_0}[/imath]
Is [imath]\aleph_0^{\aleph_0}=2^{\aleph_0}[/imath] or [imath]\aleph_0^{\aleph_0}>2^{\aleph_0}[/imath] Why? | 183513 | What is [imath]\aleph_0[/imath] powered to [imath]\aleph_0[/imath]?
By definition [imath]\aleph_1 = 2 ^{\aleph_0}[/imath]. And since [imath]2 < \aleph_0[/imath], then [imath]2^{\aleph_0} = {\aleph_1} \le \aleph_0 ^ {\aleph_0}[/imath]. However, I do not know what exactly [imath]\aleph_0 ^ {\aleph_0}[/imath] is or how I could compute it. |
316173 | Find all solutions of the equation [imath]x! + y! = z![/imath]
Not sure where to start with this one. Do we look at two cases where [imath]x<y[/imath] and where [imath]x>y[/imath] and then show that the smaller number will have the same values of the greater? What do you think? | 105821 | Finding all the numbers that fit [imath]x! + y! = z![/imath]
I have the formula [imath]x! + y! = z![/imath] and I'm looking for positive integers that make it true. Upon inspection it seems that x = y = 1 and z = 2 is the only solution. The problem is how to show it. From the definition of the factorial function we know [imath]x! = x(x-1)(x-2)...(2)(1)[/imath] So we can do something like this: [imath] [x(x-1)(x-2)...(2)(1)] + [y(y-1)(y-2)...(2)(1)] = [z(z-1)(z-2)...(2)(1)][/imath] we can then factor all of the common terms out on the LHS. [imath] [...(2)(1)][x(x-1)(x-2)... + y(y-1)(y-2)...] = [z(z-1)(z-2)...(2)(1)][/imath] and divide the common terms out of the right hand side [imath][x(x-1)(x-2)...] + [y(y-1)(y-2)...] = [z(z-1)(z-2)...][/imath] I'm stuck on how to proceed and how to make a clearer argument that there is only the one solution (if indeed there is only the one solution). If anybody can provide a hint as to how to proceed I would appreciate it. |
318080 | Prove that [imath]\frac{\int_0^1xf^2(x) \mathrm{d}x}{\int_0^1 xf(x) \mathrm{d}x}\le\frac{\int_0^1 f^2(x) \mathrm{d}x}{\int_0^1 f(x) \mathrm{d}x}[/imath]
Let [imath]f:[0,1]\rightarrow\mathbb{R_+}[/imath] be a monotone decreasing function. We want to prove that [imath]\frac{\int_0^1x(f(x))^2 \,\mathrm{d}x}{\int_0^1 xf(x) \,\mathrm{d}x}\le\frac{\int_0^1 (f(x))^2 \,\mathrm{d}x}{\int_0^1 f(x) \,\mathrm{d}x}[/imath] What would you propose here? Thanks! Sis. | 214636 | If [imath]f[/imath] is a positive, monotone decreasing function, prove that [imath]\int_0^1xf(x)^2dx \int_0^1f(x)dx\le \int_0^1f(x)^2dx \int_0^1xf(x)dx[/imath]
If [imath]f[/imath] is a positive, monotone decreasing function, prove that [imath]\int_0^1xf(x)^2dx \int_0^1f(x)dx\le \int_0^1f(x)^2dx \int_0^1xf(x)dx[/imath] |
318036 | The dimension of the intersection of two matrices ranges
I’d like to ask the following: Considering the matrices [imath]A[/imath] and [imath]B[/imath], both with the dimensions [imath] n \times m[/imath]. The ranges are [imath]R(A)[/imath] and [imath]R(B)[/imath]. I would like to find: [imath] D= \dim( R(A) \cap R(B)) [/imath] expressed in the orthonormal basis of the intersection subspace. Can you help me? I sincerely thank you! :) All the best GoodSpirit | 189285 | Calculate intersection of vector subspace by using gauss-algorithm
There are two vector subspaces in [imath]R^4[/imath]. [imath]U1 := [(3, 2, 2, 1), (3, 3, 2, 1), (2, 1, 2 ,1)][/imath], [imath]U2 := [(1, 0, 4, 0), (2, 3, 2, 3), (1, 2, 0, 2)][/imath] My idea was to calculate the Intersection of those two subspaces by putting all the given vector elements in a matrix (a vector is a column). If I run the gauss-algorithm, this leads to a matrix \begin{pmatrix} 1 & 0 & 0 & 0 & -6 & -4 \\\\ 0 & 1 & 0 & 0 & 3 & 2 \\\\ 0 & 0 & 1 & 0 &6 & 4 \\\\0 &0 & 0 & 1 & -1 & -1 \end{pmatrix} So I see that the dimension of [imath]U1 + U2[/imath] equals 4, as there are 4 linear independent vectors. Is it somehow possible to get a basis of [imath][U1] \cap [U2][/imath] from this matrix? I know that it has to be one dimensional as the dimension of [imath]U2[/imath] equals 2. |
318141 | Ideal of polynomials in [imath]k[X_1,...X_n][/imath] vanishing at a point [imath]p[/imath] is [imath](X_1 - p_1, ...,X_n - p_n)[/imath]
I'm having a little trouble following Eisenbud here: My problem is that I don't see how the isomorphism [imath]{k[x_1,...,x_n] \over \mathfrak{m}_p} \cong k[/imath] is constructed. This seems a bit hand-wavey to me. Any clarification would be gladly appreciated! Thank you. | 316769 | Points and maximal ideals in polynomial rings
Let [imath]k[/imath] be a field, then I want to prove the following statement: for every [imath]P=(b_1,\ldots,b_n)\in K^n[/imath], the ideal [imath]\mathfrak{m}_P=(x_1-b_1,\ldots,x_n-b_n)[/imath] is maximal in the polynomial ring [imath]k[x_1,\ldots,x_n][/imath]. To prove this, I consider the evaluation map [imath]v_P:k[x_1,\ldots,x_n]\longrightarrow k[/imath] sending a polynomial [imath]f(x_1,\ldots,x_n)[/imath] to [imath]f(b_1,\ldots,b_n)[/imath]. Then [imath]v_P[/imath] is a surjective morphism of rings. So we have that the quotient of [imath]k[x_1,\ldots,x_n][/imath] by the kernel of [imath]v_P[/imath] is isomorphic to [imath]k[/imath], which is a field, thus is a field itself and [imath]\ker v_P[/imath] is maximal. So we are left to prove that [imath]\mathfrak{m}_P=\ker v_P[/imath]. One of the inclusions is obvious, by definition of [imath]\mathfrak{m}_P[/imath]. On the other side, I don't know how to prove that [imath]\ker v_P[/imath] is contained in [imath]\mathfrak{m}_P[/imath]. |
286462 | On normal subgroup of a finite group
Let [imath]H[/imath] be normal subgroup of a finite group [imath]G[/imath] such that [imath](\mid H\mid, [G:H])=1[/imath]. Then prove [imath]G[/imath] has only one subgroup of order [imath]\mid H\mid[/imath]. | 169949 | If [imath]H\unlhd G[/imath] with [imath](|H|,[G:H])=1[/imath] then [imath]H[/imath] is the unique such subgroup in [imath]G[/imath].
Here is a problem from "An introduction to the Theory of Groups" by J.J.Rotman: Let [imath]G[/imath] be a finite group, and let [imath]H[/imath] be a normal subgroup with [imath](|H|,[G:H])=1[/imath]. Prove that [imath]H[/imath] is the unique such subgroup in [imath]G[/imath]. I assumed there was another normal subgroup like [imath]H[/imath], say [imath]K[/imath], such that [imath](|K|,[G:K])=1[/imath]. My aim was to show that [imath][K: K\cap H]=1 [/imath] that was not held if I didn’t suppose [imath]|H|=|K|[/imath] . My question is if my last assumption about two subgroups is right? If it isn’t, please guide me. Thanks. |
318460 | Group, normality, subgroup question.
Let [imath]G[/imath] be a group, and [imath]N,H[/imath] be subgroups of [imath]G[/imath] with [imath]N[/imath] normal. Show that [imath]HN = \{hn \mid h \in H, n \in N \}[/imath] is a subgroup of [imath]G[/imath]. Thanks in advance! | 147877 | The product of a subgroup and a normal subgroup is a subgroup
Let [imath]G[/imath] be a group, [imath]H[/imath] a subgroup of [imath]G[/imath], and [imath]N[/imath] a normal subgroup of [imath]G[/imath]. Verify that [imath]HN=\{hn\mid h \in H, n \in N\}[/imath] is a subgroup of [imath]G[/imath]. |
110459 | Is there a simple proof of the fact that if free groups [imath]F(S)[/imath] and [imath]F(S')[/imath] are isomorphic, then [imath]\operatorname{card}(S)=\operatorname{card}(S')?[/imath]
Theorem. If [imath]F(S)[/imath] and [imath]F(S')[/imath] are isomorphic free groups with bases [imath]S[/imath] and [imath]S'[/imath] respectively, then [imath]\operatorname{card}(S)=\operatorname{card}(S').[/imath] I know a proof of this fact that uses the abelianizations of the groups. It seems a bit random to me. Is there a proof that uses only basic properties of free groups? Perhaps it would be less vague to put it this way: is there a proof that a student could possibly find who has just read the definition of a free group as a set of words over an alphabet and knows that it can also be seen as a group with a certain universal property? I will be glad to accept an upvoted answer saying that no such proof is known. (And therefore doesn't exist.) | 35229 | free groups: [imath]F_X\cong F_Y\Rightarrow|X|=|Y|[/imath]
I'm reading Grillet's Abstract Algebra. Let [imath]F_X[/imath] denote the free group on the set [imath]X[/imath]. I noticed on wiki the claim [imath]F_X\cong\!\!F_Y\Leftrightarrow|X|=|Y|.[/imath] How can I prove the right implication (find a bijection [imath]f:X\rightarrow Y[/imath]), i.e. that the rank is an invariant of free groups? I am hoping for a simple and short proof, having all the tools of Grillet at hand. Rotman (Advanced Modern Algebra, p.305) proves it only for [imath]|X|<\infty[/imath], Bogopolski's (Introduction to Group Theory, p.55) proof seems (unnecessarily?) complicated, and Lyndon & Schupp's (Combinatorial Group Theory, p.1) proof I don't yet understand. It's the very first proposition in the book; in the proof, they say: The subgroup [imath]N[/imath] of [imath]F[/imath] generated by all squares of elements in [imath]F[/imath] is normal, and [imath]F/N[/imath] is an elementary abelian [imath]2[/imath]-group of rank [imath]|X|[/imath]. (If [imath]X[/imath] is finite, [imath]|F/N|=2^{|X|}[/imath] finite; if [imath]|X|[/imath] is infinite, [imath]|F/N|=|X|[/imath]). [imath]\square[/imath] Is [imath]N:=\langle w^2;w\in F\rangle[/imath]? What is an abelian [imath]2[/imath]-group? Elementary? What and how does the above quote really prove? I'm guessing a free abelian group on [imath]X[/imath] is [imath]\langle X|[X,X]\rangle\cong\bigoplus\limits_{x\in X} \mathbb{Z}[/imath]? Can an isomorphism [imath]\varphi:F_X\rightarrow F_Y[/imath] not preserve the length of words? At least one letter words? |
318727 | Bound on Permutations
I am trying to prove the following inequality, [imath]n^{(l)} = n(n-1)\cdots(n-l+1) \geq \frac{n^l}{e}\quad\text{ for }\quad 2 \leq l \leq \sqrt{n}\;.[/imath] So my approach is to observe that [imath]n^{(l)} = \dfrac{n!}{(n-l)!}[/imath] and apply Stirling's Approximation, to get [imath]n^{(l)} \geq \frac{\sqrt{2\pi n}n^ne^{-n}}{e\sqrt{n-l}(n-l)^le^{l-n}}\;.[/imath] So in order to get to the desired inequality it suffices to show that for [imath]2 \leq l \leq \sqrt{n}[/imath], [imath]\sqrt{2\pi}n^{n-l+1/2} \geq e^l(n-l)^{n-l+1/2}\;,[/imath] but I'm not sure what to do or whether my approach is optimal. It seems like this is the right approach as it would allow me to find the condition on [imath]l[/imath]. | 316279 | [imath]l^{th}[/imath] factorial bound
Show that [imath]n^{(l)}=n(n-1)\dots(n-l+1)\ge {n^l\over e}[/imath] where [imath]2\le l \le \sqrt{n}[/imath] Here is how far I've got with this: [imath]n^{(l)}=\prod_{i=0}^{l-1}(n-i)=n^l\prod_{i=1}^{l-1}(1-{i\over n})\\ \ge n^l (1-{l-1\over n})^{l-1}\ge n^l (1-{l-1\over n})^{n} [/imath] Unfortunately this is not in the right form to apply a bound on the exponential function. Also I am worried that I haven't used the constraint on [imath]l[/imath] either. Could anyone help me with this? |
319124 | Let [imath]f(z)[/imath] be an analytic function on a open connected subset [imath]\Bbb{G}[/imath] of [imath]\Bbb{C}[/imath] with [imath]|f(z)|= z_{0}[/imath] for some fixed [imath]z_{0}[/imath]
Let [imath]f(z)[/imath] be an analytic function on a open connected subset [imath]\Bbb{G}[/imath] of [imath]\Bbb{C}[/imath] with [imath]|f(z)|= z_{0}[/imath] for some fixed [imath]z_{0}[/imath] then prove that [imath]f(z)[/imath] is a constant function. | 147834 | Why: A holomorphic function with constant magnitude must be constant.
How can I prove the following assertion? Let [imath]f[/imath] be a holomorphic function such that [imath]|f|[/imath] is a constant. Then [imath]f[/imath] is constant. Edit: The more elementary the proof, the better. I'm working my way through a complex analysis workbook, and by this exercise, the only substantial thing covered has been the Cauchy-Riemann equations. |
319167 | How to find all automorphisms of [imath]\mathbb{Q}(\sqrt[3]{5})[/imath]?
Find all automorphisms of [imath]\mathbb{Q}(\sqrt[3]{5})[/imath]. How can I solve the above problem ? Please help someone. | 313637 | Calculate [imath]\mathrm{Gal}(\mathbb{Q}(\sqrt[5]{3})/\mathbb{Q})[/imath]
I'm attempting some of my first problems in solving for Galois Groups, and this one has stumped me. What I've done so far is found that [imath]\mathbb{Q}(\sqrt[5]{3})[/imath] is not a normal extension, because the minimum polynomial of [imath]\sqrt[5]{3}[/imath] is [imath]x^5-3[/imath], and [imath]\mathbb{Q}(\sqrt[5]{3})[/imath] is not the splitting field of this polynomial. So [imath]\mathbb{Q}(\sqrt[5]{3})[/imath] is not normal, thus not a Galois Extension. I know if it was a Galois extension you can determine the order of the Galois Group since it equals the degree of the extension over [imath]\mathbb{Q}[/imath]. But for an extension that is not Galois I can only get that the degree of the extension is an upper bound on the order of the Galois group. And from here I don't know where to go... Any help would be greatly appreciated! |
123796 | Algebraic extensions and polynomial evaluation
Let [imath]F\subset K[/imath] be a field extension and [imath]\alpha_1,\dots,\alpha_n\in F[/imath] algebraic elements over [imath]K[/imath]. Is it true that [imath]K(\alpha_1,\dots,\alpha_n)=K[\alpha_1,\dots,\alpha_n]\ \text{?}[/imath] Indeed, [imath]\alpha_1[/imath] is algebraic over [imath]K[/imath], thus [imath]K(\alpha_1)=K[\alpha_1][/imath]. Then, [imath]\alpha_2[/imath] is algebraic over [imath]K(\alpha_1)\subset K[/imath], thus [imath]K(\alpha_1,\alpha_2)=K(\alpha_1)(\alpha_2)=K(\alpha_1)[\alpha_2]=K[\alpha_1][\alpha_2]=K[\alpha_1,\alpha_2][/imath]. By induction, we get the result. Is there a flaw ? I don't feel sure with this. | 320000 | Algebraic numbers and fraction field.
If [imath]a\in\mathbb{C}[/imath] is algebraic, then [imath]\mathbb{Q}(a)=\mathbb{Q}[a][/imath], and the converse holds too. I am having trouble proving that the same holds for several elements. Is it true that [imath]a_1,\ldots, a_n[/imath] are algebraic iff [imath]\mathbb{Q}(a_1,\ldots, a_n)=\mathbb{Q}[a_1,\ldots, a_n][/imath]? |
320715 | convergence of weighted average. proof
It is well known that for any sequence [imath]\{x_n\}[/imath] in a normed space which converges to a limit [imath]x[/imath], the sequence of averages of the first [imath]n[/imath] terms is also convergent to [imath]x[/imath]. That is, the sequence [imath]\{a_n\}[/imath] defined by [imath]a_n = \frac{x_1+x_2+\ldots + x_n}{n}[/imath] converges to [imath]x[/imath]. Does anyone know how to prove this? Thanks. | 155839 | On Cesàro convergence: If [imath] x_n \to x [/imath] then [imath] z_n = \frac{x_1 + \dots +x_n}{n} \to x [/imath]
I have this problem I'm working on. Hints are much appreciated (I don't want complete proof): In a normed vector space, if [imath] x_n \longrightarrow x [/imath] then [imath] z_n = \frac{x_1 + \dots +x_n}{n} \longrightarrow x [/imath] I've been trying adding and subtracting inside the norm... but I don't seem to get anywhere. Thanks! |
305766 | Why is the Cartesian product of a set [imath]A[/imath] and empty set an empty set?
Let [imath]A \times \emptyset = \{(x,y)| x\in A, y \in \emptyset \}[/imath]. We know there is no element in [imath]\emptyset[/imath]. But how does it follow that [imath]A \times \emptyset = \emptyset [/imath]? | 177190 | Is the product of two sets well-defined if one is empty
Let [imath]X[/imath] be a set. What is [imath]X\times \emptyset[/imath] supposed to mean? Is it just the empty set? |
321693 | [imath]\sum \frac{1}{k}[/imath] , sum of the harmonic series
How do we sum up this series in terms of [imath]n[/imath]? [imath]1 + \frac{1}{2} + \ldots+\frac{1}{n}[/imath] Can we create a formula in terms of [imath]n[/imath] for this series? | 5035 | Is there any formula for the series [imath]1 + \frac12 + \frac13 + \cdots + \frac 1 n = ?[/imath]
Is there any formula for this series? [imath]1 + \frac12 + \frac13 + \cdots + \frac 1 n .[/imath] |
321877 | [imath]\frac{\mathrm d^n}{\mathrm d x^n} e^{-\frac {1}{x^2}} = 0[/imath] at [imath]x=0[/imath]
This is an exercise from David Brannan's Mathematical Analysis. I've proved parts (a) - (c) but need help with Part (d). Any guidance appreciated. EDIT I have solved it, by induction using the results of parts (a) and (c). | 320582 | Let [imath]g(x)=e^{-1/x^2}[/imath] for [imath]x\not=0[/imath] and [imath]g(0)=0[/imath]. Show that [imath]g^{(n)}(0)=0[/imath] for all [imath]n\in\Bbb N[/imath].
Let [imath]g(x)=e^{-1/x^2}[/imath] for [imath]x\not=0[/imath] and [imath]g(0)=0[/imath]. Show that [imath]g^{(n)}(0)=0[/imath] for all [imath]n\in\Bbb N[/imath]. In the text it is already proven that for the function [imath]f[/imath] with [imath]f(x)=e^{-\frac{1}{x}}[/imath] for [imath]x>0[/imath] and [imath]f(x)=0[/imath] for [imath]x\leq 0[/imath], we have [imath]f^{(n)}(0)=0[/imath]. This is what we thought: As [imath]g(x)=f(x^2)[/imath] for [imath]x\in\Bbb R[/imath], we may be able to use this. As [imath]g^{(n)}[/imath] becomes very hairy after taking a couple of derivatives. Any ideas ? |
322713 | If a group [imath]G[/imath] has only finitely many subgroups, then show that [imath]G[/imath] is a finite group.
If a group [imath]G[/imath] has only finitely many subgroups, then show that [imath]G[/imath] is a finite group. I have no idea on how to start this question. Can anyone guide me? | 22996 | Finite number of subgroups [imath]\Rightarrow[/imath] finite group
I'm trying to prove that any group [imath]G[/imath] of infinite order has an infinite number of subgroups. I think that if the group has an element of infinite order, then it's easy because I can take the groups generated by the powers of this element. What if it doesn't? Every element generates a cyclic subgroup. Every element belongs to at least one cyclic subgroup (that generated by itself). So the group is the union of its cyclic subgroups. If all these are finite, we would have to have an infinite collection of subgroups anyway. Is that correct? |
284297 | Maximal ideals in [imath]K[X_1,\dots,X_n][/imath]
Let [imath]K[/imath] be a field, and [imath]a_1,\dots,a_n \in K[/imath]. Prove that the ideal [imath](X_1-a_1,\dots,X_n-a_n)[/imath] is maximal in [imath]K[X_1,\dots,X_n][/imath]. I tried proving that the only elements outside the ideal are the invertibles of [imath]K[/imath] (I should still prove that this implies maximality, but it shouldn't be too difficult). Is there a better strategy, or another stategy? | 1027086 | Maximal ideal in [imath]K[x_1,...,x_n][/imath]
I'm having some difficulty with this homework problem: If [imath]A=K[x_1,...,x_n][/imath], [imath]K[/imath] a field and [imath]a_1,a_2,...,a_n [/imath] [imath]\in K[/imath]. The ideal [imath]m=<x_1-a_1,...,x_n-a_n>[/imath] is maximal. |
323321 | Invert of Matrix I-BA
Suppose [imath]A[/imath] and [imath]B[/imath] are two square Matrix. Let [imath]I-AB[/imath] be invertible. I would like to know why [imath]I-BA[/imath] is also invertible? Also what is invert of [imath]I-BA[/imath]? Thanks. | 237779 | [imath]I_m - AB[/imath] is invertible if and only if [imath]I_n - BA[/imath] is invertible.
The problem is from Algebra by Artin, Chapter 1, Miscellaneous Problems, Exercise 8. I have been trying for a long time now to solve it but I am unsuccessful. Let [imath]A,B[/imath] be [imath]m \times n[/imath] and [imath]n \times m[/imath] matrices. Prove that [imath]I_m - AB[/imath] is invertible if and only if [imath]I_n - BA[/imath] is invertible. Please provide with only a hint. |
323519 | Is [imath]\sum \frac{1}{a^{1+{1\over a}}}[/imath] convergent?
Is [imath]\sum \frac{1}{a^{1+{1\over a}}}[/imath] convergent? I know some fact that if [imath]\sum \frac{1}{a^s}[/imath] is convergent for any [imath]s>1[/imath] but here, the power is varying, more precies it is decreasing to 1 but i don't know whether it is convergent or divergent and how to prove to test or to to prove that it is divergent. | 268639 | Convergence or divergence of the series [imath]\sum_{n=1}^{\infty}\left(\frac{1}{n}\right)^{1+\frac{1}{n}}[/imath]
I have problem in determining the convergence of the series [imath]\sum_{n=1}^{\infty}\left(\frac{1}{n}\right)^{1+\frac{1}{n}}[/imath]. It seems like it is convergent given that [imath](1+\frac{1}{n})>1[/imath] for all n, but I still cannot prove it rigorously. Can anyone help me ?? |
323887 | Examples of isometries of [imath]S^2[/imath]
Hey guys I am trying to do some practice problems for my course and I came across this problem, like I know what isometries are and how they work. But I am getting confused on how I would apply the same procedure to a sphere and I am confused on that. 1) Find two rotations of the sphere [imath]r[/imath], [imath]r'[/imath] such that [imath]rr' = r'r[/imath]. 2) Find two rotations of the sphere [imath]r[/imath], [imath]r'[/imath] such that [imath]rr' \neq r'r[/imath]. 3) Find an isometry of [imath]S^2[/imath] with no fixed points. Please help out, itll be really appreciated thank you | 321984 | Give an example of two rotations that commute in [imath]S^2[/imath].
spherical geometry question involving isometries in [imath]S^2[/imath] Give an example of two rotations [imath]R_1[/imath] and [imath]R_2[/imath], of the sphere [imath]x^2 + y^2 + z^2 =1[/imath] such that [imath]R_1R_2 = R_2R_1[/imath] Also, give an example of two rotations in [imath]S^2[/imath] that do not commute. Rotation involves rotation of the sphere around an axis. |
323902 | bonus question on quaternions
I have this asterisk question, I know its hard to do and I know no one would get it in my class. Just wondering if any of you guys could give me good hints in how to do this. It would be appreciated. Thank you Question The quaternions extend the complex numbers and give non-commutative products on points of [imath]\mathbb{R^4}[/imath]. They satisfy every field axiom except for the commutativity of multipication. Show that there is no commutative product making [imath]\mathbb{R^n}[/imath] a field for [imath]n \geq 3[/imath]. For [imath]n=2[/imath] the field is the complex numbers. | 323226 | There isn't a product operation that is commmutative on [imath] \mathbb{R}^{n} [/imath] that satisfies all the field axioms for [imath] n \geq 3 [/imath].
This proof is broken down into simple easy algebra and vector questions. I would like to discuss different answers and approaches. Please see pg 162-163 on books.google.ca/books?isbn=0387290524 There are 5 questions from 7.6.3 - 7.6.7. You can read the paragraph above 7.6.3. You can also read the first part on quarternions. Exclude the "rotations of ijk space section. Here is what I have tried. Q1: I used the Pythagorean Theorem to get the norm equal to [imath] \sqrt{2} [/imath]. Then I used the property [imath] \text{Norm}(uv) = \text{Norm}(u) \text{Norm}(v) [/imath] to show that that [imath] 2 = \text{Norm}(1 - i^{2}) [/imath]. Am I right? Q2: I used the property [imath] \text{Norm}(uv) = \text{Norm}(u) \text{Norm}(v) [/imath] since [imath] \text{Norm}(i) = 1 [/imath]. But I don’t know how to show [imath] i^{2} = -1 [/imath]. One says to use the Triangle Inequality. I guess the equality implies that [imath] i^{2} [/imath] and [imath] 1 [/imath] are collinear. Q3: And thus I don’t know how to do this. Q4: The map [imath] p \longmapsto pi [/imath] multiplies all distances in [imath] \mathbb{R}^{n} [/imath] by [imath] |i| = 1 [/imath], since [imath] |pi| = |p||i| [/imath]. For any points [imath] p_{1} [/imath] and [imath] p_{2} [/imath] in [imath] \mathbb{R}^{n} [/imath], [imath] |p_{1} * i - p_{2} * i| = |(p_{1} - p_{2})i| = |p_{1} - p_{2}||i| [/imath]. Therefore, the distance [imath] |p_{1} - p_{2}| [/imath] between any two points is multiplied by [imath] |i| = 1 [/imath]. Therefore, the map is an isometry of [imath] \mathbb{R}^{n} [/imath]. Therefore, since [imath] i [/imath] and [imath] j [/imath] are perpendicular directions, [imath] i * i [/imath] and [imath] i * j [/imath] are still perpendicular by the isometry. (An isometry preserves the distance between points.) Still not sure why [imath] \mathbf{1} [/imath] and [imath] ij [/imath] are perpendicular. Q5: From [imath] jiij= j i^{2} j = jj i^{2} = j^{2} i^{2} = - \mathbf{1} * - \mathbf{1} = \mathbf{1} [/imath], therefore [imath] 1 = -1 [/imath], which is a contradiction. |
323932 | A calculus problem from Spivak
Suppose [imath]f[/imath] is integrable on [[imath]a[/imath],[imath]b[/imath]] and [imath]f(x)>0[/imath] for all [imath]x[/imath] in [[imath]a[/imath],[imath]b[/imath]]. Prove that [imath]\displaystyle\int_a^b f(x) \mathrm{d}x>0[/imath]. | 313456 | Prove integral is greater than [imath]0[/imath]
[imath]f(x)[/imath] is Riemann integrable on [imath]I=[a,b][/imath] and [imath]f(x)>0[/imath] for all [imath]x \in I[/imath], prove [imath]\int_a^b f(x) dx >0[/imath] . Need help on this question, please help me |
324109 | Probability: What Is The Average Number of Steps?
[imath]A[/imath] and [imath]B[/imath] play a game. [imath]A[/imath] has [imath]n_{A}\geq 0[/imath] dollars and player B has [imath]n_{B}\geq 0[/imath] dollars. A fair coin is tossed. If it is heads, [imath]B[/imath] gives a dollar to [imath]A[/imath]. If tails, [imath]A[/imath] gives a dollar to [imath]B[/imath]. The game stops when one of the players loses all of his/her money. What is the average number of steps until the end of the game? (Hint: Let [imath]m_{j}[/imath] be the expected number of steps required, when player [imath]A[/imath] has [imath]j[/imath] dollars and try to set up a recursive equation for [imath]m_{j}[/imath]. Find out if the recursive equation is solvable.) Thank you for your help. | 307187 | Expected number of steps till a random walk hits a or -b.
On wikipedia I read that the expected number of steps till a 1D simple random walk hits either [imath]a[/imath] or [imath]-b[/imath] is equal to [imath]ab[/imath]. (I have seen this result also on other websites.) However, no proof or further reference is given. Could someone please explain how they arrived at this result? Thanks in advance, Claus |
324183 | Polynomials generates the ideal
[imath]Hello, everyone[/imath] Let [imath]F[/imath] and field [imath]n[/imath] positive integer. Let [imath]g[/imath] element [imath]F[t][/imath] given by [imath]g (t) = t ^ {n}[/imath]. Show that [imath]g[/imath] generates the ideal [imath]J[/imath] of [imath]F [t][/imath] consisting of all polynomials of degree greater than or equal to [imath]n[/imath]. Is this statement false? For example if n=5 then, [imath]g (t) = t ^ {5}[/imath] , but g(t) doesn't generate all polynomials of degee greater or equal to n, because [imath]t^{7}+t^{6}+t^{5}-t^{4}+t^{3}-3[/imath] doesn't be generate for [imath]t^{5}[/imath] or for example [imath]t^{5}+t^{4}[/imath] neither How can I say that this statement is false? sorry for the inconvenience, but my teacher said me that the statement is true and I have a lingering doubt and this is the reason for I repeated it Thanks for you help :D Nice day. | 322058 | Polynomials and ideal
Let [imath]F[/imath] and field [imath]n[/imath] positive integer. Let [imath]g[/imath] element [imath]F[t][/imath] given by [imath]g (t) = t ^ {n}[/imath]. Show that [imath]g[/imath] generates the ideal [imath]J[/imath] of [imath]F [t][/imath] consisting of all polynomials of degree greater than or equal to [imath]n[/imath]. Could someone help me? Thanks, I am sorry for any inconvenience |
161599 | Find all primes [imath]p[/imath] such that [imath]\dfrac{(2^{p-1}-1)}{p}[/imath] is a perfect square
Find all primes [imath]p[/imath] such that [imath]\dfrac{(2^{p-1}-1)}{p}[/imath] is a perfect square. I tried brute-force method and tried to find some pattern. I got [imath]p=3,7[/imath] as solutions . Apart from these I have tried for many other primes but couldn't find any other such prime. Are these the only primes that satisfy the condition ? . If yes , how to prove it theoretically and if not, how to find others?. Thanks in advance!! | 1036541 | Find all primes p such that [imath]\frac{2^{p-1}-1}{p}[/imath] is a perfect square.
Find all primes [imath]p>2[/imath] such that [imath]\frac{2^{p-1}-1}{p}[/imath] is a perfect square. My try: Since [imath]p>2[/imath], we can write [imath]\frac{2^{p-1}-1}{p}=\frac{(2^{\frac{p-1}{2}}-1)(2^{\frac{p-1}{2}}+1)}{p}=n^2[/imath] Now since p is a prime one of the [imath](2^{\frac{p-1}{2}}-1)[/imath] and [imath](2^{\frac{p-1}{2}}+1)[/imath] case-1 [imath]p|(2^{\frac{p-1}{2}}-1)[/imath] [imath](2^{\frac{p-1}{2}}-1)=pk[/imath] hence [imath]n^2=k(pk+2)[/imath] writing [imath]k=m^2t[/imath] where [imath]t[/imath] doesn't contain any perfect square divisor [imath]a^2=t(pm^2t+2)[/imath] This indicates that t is either [imath]1[/imath] or [imath]2[/imath]. Putting [imath]t=1[/imath] [imath]a^2=pm^2+2[/imath] hence [imath]a^2=2^{\frac{p-1}{2}}+1[/imath] [imath](a-1)(a+1)=2^{\frac{p-1}{2}}[/imath] It can easily be shown that [imath]a=3[/imath] is the only solution to this equation hence [imath]p=7[/imath] is the only solution in case-1.(putting [imath]t=2[/imath] doesn't give any solution.) Similar method gives [imath]p=3[/imath] in case-2 But someone told me there is an elegent and novel method to solve it. I tried but i didn't get it. Please help. |
324837 | The set of zero-divisors in a ring [imath]R[/imath] containing maximal ideals
Let [imath]Z(R)[/imath] be the set of all zero-divisors of [imath]R[/imath]. Let [imath]p[/imath] and [imath]q[/imath] be maximal ideals contained in [imath]Z(R)[/imath] with [imath]p\cap q\neq\{0\}[/imath]. Show that [imath]Z(R)=p\cup q[/imath]. | 324789 | Is [imath]Z(R)[/imath] a maximal ideal?
If [imath]p[/imath] and [imath]q[/imath] are two maximal ideals in the set of zero-divisors in a ring [imath]R[/imath] with non-zero intersection between [imath]p[/imath] and [imath]q[/imath]. does the set of all zero-divisors are a maximal ideal and equal the union of [imath]p[/imath] and [imath]q[/imath]? Different phrasing: Let [imath]Z(R)[/imath] be the set of all zero-divisors of [imath]R[/imath]. Let [imath]p[/imath] and [imath]q[/imath] be maximal ideals contained in [imath]Z(R)[/imath] with [imath]p\cap q\neq\{0\}[/imath]. Show that [imath]Z(R)=p\cup q[/imath]. |
114497 | Vector path length of a hypotenuse
Consider the red path from A that zigzags to B, which takes [imath]n[/imath] even steps of length [imath]w[/imath]. The path length of the route [imath]P_n[/imath] will be equal to: [imath] P_n = P_x + P_y = \frac{n}{2}\times w + \frac{n}{2}\times w = n \times w [/imath] But [imath]\frac{n}{2}\times w = 1[/imath] beacuse it is the length of one of the sides of the triangle so: [imath]P_n = 2[/imath] Which will be true no matter how many steps you take. However in the limit [imath]n \to \infty, w \to 0[/imath] the parth length [imath]P_\infty[/imath] suddenly becomes: [imath]P_\infty = \sqrt{1^2 + 1^2} = \sqrt{2}[/imath] Due to Pythagoras. Why is this the case? It seems the path length suddenly decreases by 0.59! | 658277 | Ant on a square
There is a square of side [imath]1 m[/imath] and an ant has to cross diagonally. However, it chooses to walk along the boundary so the distance covered by it is [imath]2m[/imath] and not [imath]\sqrt{2} m[/imath]. This it does in two moves. Now the ant decides to walk along the one side and stop at [imath]\frac{1}{2}[/imath] that is [imath]0.5 m [/imath] and then go [imath]0.5m[/imath] up, [imath]0.5m[/imath] forward and another [imath]0.5m[/imath] upward to reach its destination. It does the same with [imath]0.25 m[/imath] ([imath]\frac{1}{4})[/imath]etc. Assume now that the ant divides the segment [imath]n[/imath] times, that is, [imath]\frac{1}{n}[/imath] and covers the same distance, by [imath]2n[/imath] moves. As [imath]n[/imath] tends to infinity, the path of the ant will be equal to the straight line path that measures [imath]\sqrt{2}m[/imath]. But then, if we consider this to be a normal limit problem as [imath]n[/imath] tends to infinity, the answer is always [imath]2[/imath]. Where is the contradiction? I have tried drawing the paths and checking them for large values of [imath]n[/imath] but I still couldn't spot the contradiction. |
325332 | How to express [imath]C[/imath] in terms of the sets [imath]A_n.[/imath]
Let [imath]f[/imath] be a real valued function on [imath]\mathbb R.[/imath] Consider the functions [imath]w_j(x)=\sup\{|f(u)-f(v)|:u,v\in[x-\frac{1}{j},x+\frac{1}{j}]\},[/imath] where [imath]j\in \mathbb Z^+[/imath] and [imath]x\in\mathbb R.[/imath] Define next, [imath]A_{j,n}=\{x\in\mathbb R:w_j(x)<\frac{1}{n}\},n=1,2,\cdots[/imath] and [imath]A_n=\cup_{j=1}^{\infty}A_{j,n},n=1,2,\cdots[/imath] Now let [imath]C=\{x\in\mathbb R:f[/imath] is continuous at [imath]x\}.[/imath] How can we express [imath]C[/imath] in terms of the sets [imath]A_n[/imath]? | 268558 | Express [imath]C[/imath] in terms of the sets [imath]A_n[/imath]
Let [imath]f[/imath] be a real valued function on [imath]\mathbb{R}[/imath]. Consider the functions [imath] w_j(x) := \sup\left\{\vert f(u) − f (v)\vert : u,v \in \left[x−\frac{1}{j},x+\frac{1}{j}\right]\right\} [/imath] where [imath]j[/imath] is a positive integer and [imath]x\in\mathbb{R}[/imath]. Then define [imath] A_{j,n}: = \left\{x \in \mathbb{R}: w_j(x)<\frac{1}{n}\right\},\qquad n=1,2,\ldots [/imath] and [imath] A_n:=\bigcup_{j=1}^\infty A_{j,n}, \qquad n = 1,2,\ldots [/imath] Now let [imath]C=\{x \in \mathbb{R}:f \text{ is continuous at }x \}[/imath]. Express [imath]C[/imath] in terms of the sets [imath]A_n[/imath]. I am totally confused. Please help me. |
325236 | Is there a bijection between [imath]\mathbb N[/imath] and [imath]\mathbb N^2[/imath]?
Is there a bijection between [imath]\mathbb N[/imath] and [imath]\mathbb N^2[/imath]? If I can show [imath]\mathbb N^2[/imath] is equipotent to [imath]\mathbb N[/imath], I can show that [imath]\mathbb Q[/imath] is countable. Please help. Thanks, | 187751 | Cardinality of the set of all pairs of integers
The set [imath]S[/imath] of all pairs of integers can be represented as [imath]\{i \ | \ i \in \mathbb{Z} \} \times \{j\ | \ j \in \mathbb{Z}\}[/imath]. In other words, all coordinates on the cartesian plane where [imath]x, y[/imath] are integers. I also know that a set is countable when [imath]|S|\leq |\mathbb{N}^+|[/imath]. I attempted to map out a bijective function, [imath]f : \mathbb{N}^+ \rightarrow S[/imath]. [imath]1 \rightarrow (1,1) \\ 2 \rightarrow (1,2)\\ 3 \rightarrow (1,3) \\ \quad \vdots [/imath] I determined from this that the natural numbers can only keep up with [imath](1,*)[/imath]. But there is the ordered pairs where [imath]x=2,3,4,\cdots[/imath] not to mention the negative integers. In other words, [imath]|S|\geq |\mathbb{N}^+|[/imath] and therefore [imath]S[/imath] is not countably infinite. Is this correct? (I don't think it is... Something to do with my understanding of infinite sets) |
325341 | If [imath]T:V\to W[/imath] is an injective linear map, show there is a linear map [imath]L:W\to V[/imath] such that [imath]LT=1_V[/imath]
Let [imath]V[/imath] and [imath]W[/imath] be finite dimensional vector spaces over a field [imath]F[/imath]. Let [imath]T:V\to W[/imath] be a linear transformation. Suppose that [imath]T[/imath] is one-to-one. Show that there is a linear transformation [imath]L:W\to V[/imath] such that [imath]LT=1_V[/imath]. So far I have: Since [imath]T[/imath] is injective, then a basis [imath]\{v_1,\ldots,v_n\}[/imath] for [imath]V[/imath] gets sent to a linearly independent set [imath]\{T(v_1),\ldots,T(v_n)\}[/imath] of [imath]W[/imath]. I'm not sure how to expand the basis for [imath]W[/imath]. Is it just [imath]B_W=\{w_1,\ldots,w_m\}[/imath]? I know that if [imath]T[/imath] is 1-1, then [imath]\dim V\leq\dim W[/imath]. So [imath]n\leq\dim W[/imath]. Any help from here would be great. | 325197 | Linear Algebra - Transformations
Let [imath]V[/imath] and [imath]W[/imath] be finite dimensional vector spaces over a field [imath]F[/imath]. Let [imath]T:V\to W[/imath] be a linear transformation. Suppose that [imath]T[/imath] is one-to-one. Show that there is a linear transformation [imath]L:W\to V[/imath] such that [imath]LT=1_V[/imath]. The second question is Let [imath]T:V\to W[/imath] and [imath]L:W\to V[/imath] be a linear transformation. Show: [imath]T[/imath] is injective if [imath]LT=1_V[/imath] and [imath]T[/imath] is surjective if [imath]TL=1_W[/imath] I know that if 1 and 2 are true, that it is a isomophism, no idea how to prove it though. |
325556 | how to prove that there don't exist any [imath]n\in \Bbb N[/imath] such that [imath]\phi (n)=14[/imath]?
How to prove that there don't exist any [imath]n\in \Bbb N[/imath] such that [imath]\phi (n)=14[/imath] ? We know that [imath]\phi (n)= {p_1}^{\alpha_1}{p_2}^{\alpha_2}...{p_n}^{\alpha_n}(1-\frac{1}{p_1})...(1-\frac{1}{p_n}) [/imath] if [imath]n=3[/imath] then [imath]\phi (n)=2[/imath] also we have [imath]\phi (9)=6[/imath] and [imath] \phi(n_1)\phi(n_2)=\phi(n_1n_2)[/imath]. So the problem turns out to be : Does there exist [imath]n \in \Bbb N[/imath] such that [imath]\phi(n)=7[/imath] ? | 5661 | Generalizing values which Euler's-totient function does not take
I was reading about Euler's totient function on wikipedia, and it eventually led me to this book on google: Page 74 of the book, Prime numbers: the most mysterious figures in math By David G. Wells. Anyway, the books lists many assertions without proof, only references which I can't find. One I could not solve for myself said that not all even numbers are values of [imath]\phi(n)[/imath]. The sequence of even non-values of [imath]\phi(n)[/imath] starts: [imath]14, 26, 34, 38, 50, 62,\dots[/imath] After thinking about it for a while, I've made little headway. Looking at 14 specifically, I suppose if such a solution [imath]a[/imath] to [imath]\phi(x)=14[/imath] were to exist, [imath](a,m_i)=1[/imath] for [imath]1\leq i\leq 14[/imath], for some [imath]m_i\lt a[/imath] and so there must also exist inverses [imath]\overline{a_i}[/imath] such that [imath]\overline{a_i}a\equiv 1 \pmod {m_i}[/imath] for each [imath]i[/imath]. I'm looking for some contradiction, possibly of the Chinese Remainder Theorem, to show such [imath]a[/imath] cannot exist. Is there some way to generalize which values are not taken by [imath]\phi[/imath], or at least explain why this is the case? I was hoping to see why [imath]14[/imath] is the least such integer such that this is true, but values [imath]1[/imath] to [imath]13[/imath] must be indeed taken. I suppose this would also explain why 26 is the next such value that is not taken, while [imath]15[/imath] to [imath]25[/imath] are. |
325583 | Proving [imath]\left|\frac{\left(\log\left(n\right)\right)^{c}}{n^{d}}\right|[/imath] tends to zero
everyone, I'm trying to show that: [imath]\forall\, c\in\mathbb{R},\, d>0\,:\,\lim_{n\to\infty}\frac{\left(\log\left(n\right)\right)^{c}}{n^{d}}=0[/imath] Intuitively it's completely clear but formalizing it is proving to be a bit difficult. I'd appreciate some help evaluating [imath]\left|\frac{\left(\log\left(n\right)\right)^{c}}{n^{d}}\right|[/imath] in a way that would lead to the result. Thanks a lot. | 303933 | Evaluate [imath]\lim_{n\to\infty} \log^a(n)/n^b[/imath]
How can I prove that [imath] \lim_{n\to\infty} \frac{(\ln(n))^a}{n^b} = 0 \;\forall a,b > 0 [/imath] ? Intuitively it is clear to me because of the behavior of the functions. Thanks for all. Edit I'm not able to use L'Hopital rule. Sorry. |
325805 | Prove that there are [imath]736[/imath] [imath]2 \times 2[/imath] matrices ([imath]A[/imath]) where [imath]A=A^{-1}[/imath]
I'm doing some assignments to teach myself cryptology. I am still at the introductory cryptology level, where a lot of it is discrete mathematics, so I believe - and hope - that it is a somewhat simple question for someone in here. It has shown to not be that for me, as it has been holding me prisoner the whole evening and afternoon. The assignment that is bugging me, contains the following text: Let A be an invertible [imath]m \times m[/imath] matrix over [imath]\mathbb{Z}_{26}[/imath], where [imath]A=A^{-1}[/imath] and from which it follows that [imath]det(A) \equiv \pm 1[/imath] mod [imath]26[/imath]. Show that there are [imath]736[/imath] [imath]2 \times 2[/imath] matrices [imath]A[/imath] over [imath]\mathbb{Z}_{26}[/imath] for which it holds that [imath]A=A^{-1}[/imath]. Hint: Split the problem into two cases. Consider the problem first modulo [imath]2[/imath], then modulo [imath]13[/imath], and finally use the Chinese Remainder Theorem to get the result modulo [imath]26[/imath]. What I have tried to do: A lot that have not yielded any results yet... The point where I am at now, is after I have split [imath]det(A) \equiv \pm 1[/imath] mod 26 into: [imath]det(A) \equiv \pm 1[/imath] mod [imath]13[/imath] and [imath]det(A) \equiv \pm 1[/imath] mod [imath]2[/imath]. Afterwards, I have been thinking that since [imath]A=A^{-1}[/imath] means that [imath]A^2=I[/imath], what I need to do is find all cases where: [imath]\pmatrix{a&b\\c&d}^2=\pmatrix{1&0\\0&1}[/imath] Which means I have the congruences: \begin{align} a^2+bc&\equiv1\\ ab+bd&\equiv0\\ ca+dc&\equiv0\\ cb+d^2&\equiv1\ \end{align} I have no clue where to go from here though.... Or if what I have been doing so far, is the correct way of doing it, if I want to follow that hint... I hope someone can and will help out, and are willing to shed some light on this. Thanks (a lot) in advance. | 324542 | Why are there [imath]736[/imath] matrices [imath]M\in \mathcal M_2(\mathbb{Z}_{26})[/imath] for which it holds that [imath]M=M^{-1}[/imath]?
I'm currently trying to introduce myself to cryptography. I'm reading about the Hill Cipher currently in the book Applied Abstract Algebra. The Hill Cipher uses an invertible matrix [imath]M[/imath] for encryption and the inverse matrix [imath]M^{-1}[/imath] for decryption. The book recommends to use a matrix [imath]M[/imath] for which it holds that [imath]M=M^{-1}[/imath], since we then have the same key for encryption and decryption. In the book, they use a [imath]2 \times 2[/imath] matrix [imath]M[/imath] over [imath]\mathbb{Z}_{26}[/imath] as example, and state that for there are 736 [imath]2 \times 2[/imath] matrices for which it hold that [imath]M=M^{-1}[/imath]. I'm trying to pick up on as much as possible when reading things, since I find it counter-productive for learning to skip something, when you don't get the theory behind it. Can someone enlighten to me, as to why it is that there are 736 possible [imath]2 \times 2[/imath] [imath]M=M^{-1}[/imath] matrices and how to find them? |
325852 | Exactly [imath]\phi(\phi(n))[/imath] primitive roots modulo [imath]n[/imath]
If [imath]n\in \Bbb N[/imath] is a primitive root then how to prove that: There are exactly [imath]\phi(\phi(n))[/imath] primitive roots modulo [imath]n[/imath] such that are peer to peer [imath]\not \equiv[/imath] | 166866 | Prove if [imath]n[/imath] has a primitive root, then it has exactly [imath]\phi(\phi(n))[/imath] of them
Prove if [imath]n[/imath] has a primitive root, then it has exactly [imath]\phi(\phi(n))[/imath] of them. Let [imath]a[/imath] be the primitive root then I know other primitive roots will be among [imath]\{a,a^2,a^3 \cdots\cdots a^{\phi(n)} \}[/imath] because any other number will be congruent modulo [imath]n[/imath] to one of these. Then I figured as the answer is [imath]\phi(\phi(n))[/imath] so for [imath]a^k[/imath] to be primitive root, [imath]k[/imath] must be coprime with [imath]\phi(n)[/imath].But I don't know the reason.Am I missing any silly point? |
325816 | If [imath]L\rightarrow M\rightarrow N[/imath] is a short exact sequence and L and N are finite over A, then so is M
If [imath]L\rightarrow M\rightarrow N[/imath] is a short exact sequence of A-modules and [imath]L[/imath] and [imath]N[/imath] are finite over A, then so is M. | 234753 | Finitely generated modules in exact sequence
For [imath]A[/imath]-modules and homomorphisms [imath]0\to M'\stackrel{u}{\to}M\stackrel{v}{\to}M''\to 0[/imath] is exact. Prove if [imath]M'[/imath] and [imath]M''[/imath] are fintely generated then [imath]M[/imath] is finitely generated. |
326190 | If [imath]a \in \mathbb{R}[/imath] and [imath]S = \{q \in\mathbb{Q}:q, then \sup S=a[/imath]
Let [imath]a \in \mathbb{R}[/imath] and [imath]S = \{q \in\mathbb{Q}:q<a\}[/imath]. Show that [imath]\sup S=a[/imath]. | 46643 | supremum and infimum
If [imath]x \in \mathbb{R}[/imath], prove that [imath]x = \sup \{r \in \mathbb{Q}: r < x \} = \inf \{s \in \mathbb{Q}: x <s \}[/imath]. For convenience, let [imath]A = \{r \in \mathbb{Q}: r < x \}[/imath] and [imath]B = \{s \in \mathbb{Q}: x <s \}[/imath]. I think both of these sets are non-empty by the denseness of the rationals. The elements of [imath]B[/imath] are upper bounds of [imath]A[/imath], and the elements of [imath]A[/imath] are lower bounds of [imath]B[/imath]. Hence [imath]\sup A[/imath] and [imath]\inf B[/imath] exist. To actually show that these are equal to [imath]x[/imath], would you suppose [imath]x_1[/imath] was an arbitrary upper bound and show that [imath]x_1 > x[/imath] (similar thing with lower bounds)? |
326287 | Is the union of two cartesian products equal to the product of their unions?
Can we prove that [imath](A \times B)\cup (C \times D) = (A \cup C) \times (B \cup D) \;?[/imath] If my understanding is correct, we cannot prove because we do not know if [imath]A \times B[/imath] and [imath]C \times D[/imath] share common elements or not. Please tell me if I'm wrong. | 283777 | Is it always true that [imath](A_1 \cup A_2) \times (B_1 \cup B_2)=(A_1\times B_1) \cup (A_2 \times B_2)[/imath]
Is it always true that [imath](A_1 \cup A_2) \times (B_1 \cup B_2)=(A_1\times B_1) \cup (A_2 \times B_2)[/imath]? I don't believe this is true. I have tried to draw pictures to help me get on the right path, but I think that the union makes this untrue. for example, if [imath]a \in A_1[/imath] and [imath]b \in B_2[/imath], then [imath](a,b)[/imath] would not be in [imath](A_1\times B_1) \cup (A_2 \times B_2)[/imath]. Is this a correct assumption? |
326744 | Question about SO(3) and "infinitesimal generators"
I've been struggling with a paragraph appearing in this book: http://www.math.sunysb.edu/~kirillov/liegroups/liegroups.pdf ( Example 3.10 bottom half of page 31 ). It says that "by theorem 3.7, elements of the form [imath] exp(tJ_{x}), exp(tJ_{y}), exp(tJ_{z}) [/imath] generate SO(3). Where [imath] J_{x}, J_{y}, J_{z} [/imath] are the usual basis vectors for the Lie Algebra so(3) What I am struggling with is this: First of all, I know that these elements do generate SO(3), this is sometimes called the Tait-Bryan angle parametrization of SO(3). What I don't understand, is how elements of the above form generate a neighbourhood of the identity, and how it follows from anything that the book has said. I understand that he is trying to say, if they do generate a neighbourhood of the identity, then they will in turn generate the entire group, as SO(3) is connected. However, from the results of the book up to this section, I can only justify as much as saying every element in SO(3) is of the form [imath] exp(tJ) [/imath] for some J in the Lie algebra (not necessarily one of the basis elements) This is because the exponential map for a compact and connected Lie group is surjective. This is result is also known as Euler's axis-angle parametrization of SO(3). So, how is the book supposed to justify the statement: "by Theorem 3.7, elements of the form [imath] exp(tJ_{x}), exp(tJ_{y}), exp(tJ_{z}) [/imath] generate a neighbourhood of the identity"? EDIT: Sorry, Theorem 3.7 is on the page before ( page 30 ). It gives properties of the exponential map ( it is a local diffeomorphism, commutes with lie group homomorphisms..et c ) I even tried to work out the BCH formula for the product of two exponentials, but still, it is not been clear to me. Thank you | 147056 | Is a basis for the Lie algebra of a Lie group also a set of infinitesimal generators for the Lie group?
Let [imath]G[/imath] be a (EDIT: connected) Lie group of dimension [imath]n[/imath], and let [imath]\mathfrak{g}[/imath] be the associated Lie algebra. If [imath]x_1,\ldots,x_n[/imath] is a basis for [imath]\mathfrak{g},[/imath] is it necessarily true that the 1-parameter subgroups [imath]e^{tx_1},\ldots,e^{tx_n}[/imath] generate [imath]G[/imath]? Note: It is sufficient to show that the subgroup generated by the 1-parameter subgroups is closed, since it follows that it is a Lie subgroup of dimension [imath]n[/imath]. In particular, it must contain a neighborhood of the identity, which generates [imath]G[/imath]. Also, note that it is not always true that the subgroup generated by 1-parameter subgroups is a Lie subgroup; consider the 1-parameter subgroup of the 2-torus that is a line with irrational slope. |
327284 | Fourier Transform of a function under an arbitrary coordinate transform
Consider a function [imath]f(x)[/imath] and its Fourier Transform [imath]\tilde{f}(k)[/imath] given by [imath] \tilde{f}(k) = \int_\mathbb{R}\!\!\!dx\; e^{-ikx}f(x). [/imath] Now, lets have the coordinate transform [imath]\xi = \tau(x)[/imath] and, thus, we have the Fourier Transform [imath]\tilde{f}(\kappa)[/imath] of [imath]f(\xi)=f(\tau(x))[/imath] with a new coordinate [imath]\kappa[/imath]. Is there a way to compute [imath]\tilde{f}(\kappa)[/imath] from a given [imath]\tilde{f}(k)[/imath] and coordinate transform [imath]\xi = \tau(x)[/imath]? Does the coordinate transform [imath]k \rightarrow \kappa[/imath] exist at all? Thanks and regards. | 171462 | Fourier transform of function composition
Given two functions [imath]f[/imath] and [imath]g[/imath], is there a formula for the Fourier transform of [imath]f \circ g[/imath] in terms of the Fourier transforms of [imath]f[/imath] and [imath]g[/imath] individually? I know you can do this for the sum, the product and the convolution of two functions. But I haven't seen a formula for the composition of two functions. |
327681 | limit problem: [imath]\lim_{n\to +\infty}n\int^1_0 x^nf(x)dx [/imath]
Let [imath]f:[0,1]\to\mathbb R[/imath] be a continuous function.I want to calculate this limit:[imath]\lim_{n\to +\infty}n\int^1_0 x^nf(x)dx [/imath] | 225550 | Evaluate [imath]\lim_{n\to\infty}n\int_0^1x^nf(x)dx[/imath]
Let [imath]f:[0,1]\mapsto\mathbb{R}[/imath] be a continuous function. Evaluate [imath]\lim_{n\to\infty}n\int_0^1x^nf(x)dx[/imath] |
328047 | A question about elementary combinatorics
Without computing directly, how to show that [imath]\dfrac{\binom{100}{50}}{2^{100}}<0.1[/imath] easily? | 274753 | Solve this inequality [imath]\prod_{i=1}^{50} \frac {2i-1}{2i} < \frac {1}{10}[/imath]
Prove that [imath] \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \frac{7}{8}\cdot \frac{9}{10}\cdot \frac{11}{12}\cdot \frac{13}{14}...\cdot \frac{91}{92}\cdot \frac{93}{94}\cdot \frac{95}{96}\frac{97}{98}\cdot \frac{99}{100} <\frac{1}{10}[/imath] |
328020 | Conntectedness of SpecR
Let [imath]R[/imath] be a commutative ring with unity, [imath]e\in R[/imath] is called and idempotent if [imath]e^2=e[/imath] and if [imath]e\notin \{0,1\}[/imath] then it is called a non-trivial idempotent.want to show that [imath]\text{Spec}R[/imath] is not connected if and only if there exists a non trivial idempotent [imath]e\in R[/imath]. I have no idea how to solve this, need your help, thank you. | 326452 | If [imath]\mathop{\mathrm{Spec}}A[/imath] is not connected then there is a nontrivial idempotent
I'm solving a problem from Atiyah-Macdonald. I have to show that if [imath]X=\mathop{\mathrm{Spec}}A[/imath] is not connected then [imath]A[/imath] contains idempotents [imath]e \neq 0,1[/imath]. The converse is easy. If [imath]e \in A[/imath] is an idempotent then [imath](e)+(1-e)=(1)[/imath] and [imath](e)\cdot(1-e)=0[/imath] so that [imath] V(e) \cup V(1-e) = V( (e) \cdot(1-e))=V(0) = X, \\ V(e) \cap V(1-e) = V( (e)+(1-e))=V(1)=\varnothing [/imath] then [imath]V(e)[/imath] and [imath]V(1-e)[/imath] are both closed and open and [imath]X[/imath] is not connected. Now let [imath]\mathfrak{a}[/imath] and [imath]\mathfrak{b}[/imath] be ideals in [imath]A[/imath] such that [imath]V(\mathfrak{a}) \cup V(\mathfrak{b})=X[/imath], [imath]V(\mathfrak{a}) \cap V(\mathfrak{b}) = \varnothing[/imath]. Then [imath] V(\mathfrak{a}) \cup V(\mathfrak{b}) = V( \mathfrak{a} \cap \mathfrak{b} ) = X, [/imath] i.e. [imath]\left\{ \mathfrak{p} - \text{prime} \mid \mathfrak{a} \cap \mathfrak{b} \subseteq \mathfrak{p} \right\} = X[/imath], i.e. [imath]\mathfrak{a} \cap \mathfrak{b} \subseteq \cap \mathfrak{p} = \mathfrak{n}[/imath] (nilradical). On the other hand since [imath] V(\mathfrak{a}) \cap V(\mathfrak{b}) = V(\mathfrak{a}+\mathfrak{b})=\varnothing [/imath] we have [imath]\left\{ \mathfrak{p} - \text{prime} \mid \mathfrak{a}+\mathfrak{b} \subseteq \mathfrak{p} \right\} = \varnothing[/imath]. Then [imath]\mathfrak{a}+\mathfrak{b}=(1)[/imath] because any ideal that is not equal to [imath](1)[/imath] is contained in some maximal ideal. Then [imath]\mathfrak{a}[/imath] and [imath]\mathfrak{b}[/imath] are comprime and [imath]\mathfrak{a} \cdot \mathfrak{b} = \mathfrak{a} \cap \mathfrak{b}[/imath]. So I have two ideals [imath]\mathfrak{a}[/imath] and [imath]\mathfrak{b}[/imath] with properties [imath] \mathfrak{a} + \mathfrak{b} = (1), \\ \mathfrak{a} \cdot \mathfrak{b} = \mathfrak{a} \cap \mathfrak{b} = \mathfrak{n}. [/imath] I don't see any way to obtain a nontrivial idempotent [imath]e \in A[/imath] here. Please help me. |
328309 | Calculate [imath]\operatorname{Hom}_{\Bbb Z}(\Bbb Z_6,\Bbb R^*\oplus \Bbb C^*)[/imath]
Let [imath]\Bbb R^*=\Bbb R-\{0\}[/imath] (non-zero real numbers) and [imath]\Bbb C^*=\Bbb C-\{0\}[/imath] (non-zero complex numbers) be multiplicative groups. Is this equality true? [imath]\operatorname{Hom}_{\Bbb Z}(\Bbb Z_6,\Bbb R^*\oplus \Bbb C^*)=0.[/imath] We know that [imath]\operatorname{Hom}_{\Bbb Z}(\Bbb Z_6,\Bbb R^*\oplus \Bbb C^*)\underset{\Bbb Z}{\cong} \operatorname{Hom}_{\Bbb Z}(\Bbb Z_6,\Bbb R^*)\oplus \operatorname{Hom}_{\Bbb Z}(\Bbb Z_6, \Bbb C^*)[/imath] as [imath]\Bbb Z[/imath]-modules. | 327710 | Problem on [imath]\operatorname{Hom}(\mathbb Z_6,R^*\oplus C^*)[/imath]
If [imath]\mathbb R^*[/imath] and [imath]\mathbb C^*[/imath] be respectively multiplicative groups of non-zero real and complex number then whether [imath]\operatorname{Hom}(\mathbb Z_6,R^*\oplus C^*)[/imath] is isomorphic to [imath]\mathbb Z_{12}[/imath]? or [imath]\mathbb Z_6[/imath]? or [imath]\mathbb Z_6\oplus \mathbb Z_2[/imath]? |
328428 | Show that [imath]\displaystyle \sum_{n\geq{1}} ||x_n -x_{n+1}|| < \infty[/imath] is a Cauchy sequence.
If a sequence [imath](x_n)^{\infty}_{n=1}[/imath] in [imath]\mathbb{R}^n[/imath] satisfies [imath]\displaystyle \sum_{n\geq{1}} ||x_n -x_{n+1}|| < \infty[/imath], Show that it is a Cauchy sequence. Thoughts: By definition, a sequence [imath]x_k[/imath] in [imath]\mathbb{R}^n[/imath] is Cauchy if for every [imath]\varepsilon >0[/imath], there is an integer [imath]N[/imath] such that [imath]||x_k-x_l||<\varepsilon[/imath] for all [imath]k,l\geq{N}[/imath] A set [imath]S \subset {\mathbb{R}^n}[/imath] is complete if every Cauchy sequence of points in [imath]S[/imath] converges to a point in [imath]S[/imath] So, in this question, every two consecutive elements will create a new [imath]\varepsilon[/imath], say [imath](\varepsilon_1, ...., \varepsilon_n)[/imath] [imath]||x_n-x_{n+1}||<\varepsilon_1[/imath], [imath]||x_{n+1}-x_{n+2}||<\varepsilon_2[/imath], [imath]||x_{n+2}-x_{n+3}||<\varepsilon_3[/imath], ....... we have [imath]\varepsilon_1 > \varepsilon_2 > \varepsilon_3 > ....[/imath] Sum up all the [imath]\varepsilon[/imath], we must get an exact number because the sum wont get infinitely large as its getting smaller and smaller with always [imath]>0[/imath] | 208559 | Cauchy Sequence. What is this question actually telling me?
Let [imath](a_n)[/imath] be a sequence such that [imath]\lim\limits_{N\to\infty} \sum_{n=1}^n |a_n-a_{n+1}|<\infty[/imath]. Show that [imath](a_n)[/imath] is Cauchy. So basically I am told that the sum of the difference isn't infinite. I know that to show the sequence is Cauchy, the difference between the sums must be very small ([imath]\epsilon[/imath]). So what exactly do I have to do to answer this question? I am not having a good understanding what "new" information is giving me |
328552 | How to evaluate [imath]\int_{0}^{\pi }\frac{x\sin x}{1+\cos^{2}x}dx[/imath]
How can I evaluate this integral? [imath]\int_{0}^{\pi }\frac{x\sin x}{1+\cos^{2}x}dx[/imath] | 323109 | Integrating [imath]\int^0_\pi \frac{x \sin x}{1+\cos^2 x}[/imath]
Could someone help with the following integration: [imath]\int^0_\pi \frac{x \sin x}{1+\cos^2 x}[/imath] So far I have done the following, but I am stuck: I denoted [imath] y=-\cos x [/imath] then: [imath]\begin{align*}&\int^{1}_{-1} \frac{\arccos(-y) \sin x}{1+y^2}\frac{\mathrm dy}{\sin x}\\&= \arccos(-1) \arctan 1+\arccos 1 \arctan(-1) - \int^1_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\\&=\frac{\pi^2}{4}-\int^{1}_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\end{align*}[/imath] Then I am really stuck. Could someone help me? |
329369 | Pumping Lemma to show that a language is not Context Free
I am trying to use the Pumping Lemma to prove that the following language is not context free: [imath]\{0^n\mid \text{$n$ is prime}\}[/imath] I am having a really difficult time with Pumping Lemma. Up until now I was proving that a language is not regular using the Pumping lemma, but I am not sure how to begin to use the pumping lemma to prove that a language is not Contex-Free. I appreciate any suggestions, Many thanks in advance! | 217741 | How do I show language of prime number of xs not context-free?
I have a hunch that the language [imath]L = \{ x^n : n \text{ is prime.} \}[/imath] is not context-free. I am trying to show that by contradiction with the Pumping Lemma: First assume that [imath]L[/imath] is context-free. That means for any string in [imath]L[/imath] of a certain pumping length [imath]p[/imath] or greater, that string can be broken into [imath]s = uvxyz[/imath] where [imath]|vxy| \le p[/imath], [imath]|vy| > 0[/imath], and [imath]uv^ixy^iz[/imath] is in [imath]L[/imath] where [imath]i[/imath] can be any natural number including 0. I first tried letting [imath]s = x^P[/imath]. However, I'm not quite sure how to divide this value up into [imath]uxvyz[/imath] to show that it cannot be pumped. Any advice? This is not homework. I am practicing on my own. Thanks! |
329510 | Number of solutions for [imath]\displaystyle\sum\limits_{i = 1}^k a_i = n[/imath]
I am aware that number of solutions for [imath]a_1,a_2,...,a_k[/imath] for [imath]a_1+a_2+...+a_k = n[/imath] is [imath]\binom{n+k-1}{n}[/imath] For [imath]n = 3[/imath] or [imath]n = 2[/imath], it is easy to make cases and follow the result. But How to derive this result for arbitrary n and k? | 145490 | partition a number N into K tuples
Given an integer [imath]N ≥ 0[/imath] and an integer [imath]K ≥ 0[/imath], how many tuples [imath](n_1,\ldots,n_k)[/imath] are there such that [imath]n_i ≥0[/imath] and [imath]\Sigma n_i = N[/imath]? In other words, how many way can you "partition" [imath]N[/imath] into [imath]K[/imath] sets such that the sum of the sets adds up to [imath]N[/imath]. For example, [imath]n = 5[/imath], [imath]k =2[/imath] gives 6 tuples: [imath](5,0) , (4,1) \ldots (0,5)[/imath]. I've come up with this recursive definition: [imath]P(n,k) =[/imath] # of tuples [imath](n_1,..,n_k)[/imath] such that [imath]n_i ≥ 0[/imath] and [imath]\Sigma n_i = N[/imath] [imath]P(n,k) = \Sigma_{i=0}^n{P(i,k-1)}[/imath] [imath]P(0,k)=1 [/imath] [imath]P(n,1)=1[/imath] A suggested answer below is that this problem is the Stirling #'s of 2nd kind but I am not sure how to reduce this problem to it. See my response below. My question is if there's a closed form solution for [imath]P(n,k)[/imath]? |
329518 | Suppose that [imath]x[/imath] is a fixed nonnegative real number such that for all positive real numbers [imath]\epsilon[/imath], [imath]0≤x≤\epsilon[/imath]. Show that [imath]x=0[/imath].
Suppose that [imath]x[/imath] is a fixed nonnegative real number such that for all positive real numbers [imath]\epsilon[/imath], [imath]0≤x≤\epsilon[/imath]. Show that [imath]x=0[/imath]. | 321456 | Suppose that [imath]x[/imath] is a fixed nonnegative real number such that for all positive real numbers [imath]E[/imath] , [imath]0\leq x\leq E[/imath]. Show that [imath]x=0[/imath].
Suppose that x is a fixed nonnegative real number such that for all positive real numbers [imath]E[/imath] , [imath]0\leq x\leq E[/imath]. Show that [imath]x=0[/imath]. |
329621 | If a ring element is right-invertible, but not left-invertible, then it has infinitely many right-inverses.
Let [imath]A[/imath] be a ring and [imath]a\in A[/imath] an element that has a right-inverse but does not have a left-inverse. Show that [imath]a[/imath] has infinitely many right-inverses. | 156238 | Kaplansky's theorem of infinitely many right inverses in monoids?
There's a theorem of Kaplansky that states that if an element [imath]u[/imath] of a ring has more than one right inverse, then it in fact has infinitely many. I could prove this by assuming [imath]v[/imath] is a right inverse, and then showing that the elements [imath]v+(1-vu)u^n[/imath] are right inverses for all [imath]n[/imath] and distinct. To see they're distinct, I suppose [imath]v+(1-vu)u^n=v+(1-vu)u^m[/imath] for distinct [imath]n[/imath] and [imath]m[/imath]. I suppose [imath]n>m[/imath]. Since [imath]u[/imath] is cancellable on the right, this implies [imath](1-vu)u^{n-m}=1-vu[/imath]. Then [imath](1-vu)u^{n-m-1}u+vu=((1-vu)u^{n-m-1}+v)u=1[/imath], so [imath]u[/imath] has a left inverse, but then [imath]u[/imath] would be a unit, and hence have only one right inverse. Does the same theorem hold in monoids, or is there some counterexample? |
329784 | Proving [imath]\int_{0}^{\pi/2}(\log(\sin x))^2dx = \frac{1}{24}\cdot(\pi^3 + 12\pi(\log 2)^2)[/imath]
Proving [imath]\int_{0}^{\pi/2}(\log(\sin x))^2dx = \frac{1}{24}\cdot(\pi^3 + 12\pi(\log 2)^2)[/imath] without making use of gamma function,digamma function, hypergeometric function. | 300061 | A log improper integral
Evaluate : [imath]\int_0^{\frac{\pi}{2}}\ln ^2\left(\cos ^2x\right)\text{d}x[/imath] I found it can be simplified to [imath]\int_0^{\frac{\pi}{2}}4\ln ^2\left(\cos x\right)\text{d}x[/imath] I found the exact value in the table of integrals: [imath]2\pi\left(\ln ^22+\frac{\pi ^2}{12}\right)[/imath] Anyone knows how to evaluate this? |
330200 | How can one prove that integral can not be solved in transcendental functions?
For people who do not have time to read all this: Is there a simple proof (restricted to knowing calculus) that the integral can not be solved in transcendental function. Longer version: I am curious how people were able to understand that some particular integral (for example [imath]\int e^{x^2}[/imath]) can not be solved in transcendental functions? In my opinion it is so counter intuitive, and after knowing that almost every function can be differentiated (I am aware that not all), and that for every integral there exist a function that should be anti-derivative I would assume that the integral of everything should be a transcendental function. All in all, if people where trying for long time to find [imath]\int e^{x^2}[/imath] and was not able to find a transcendental function, it does not mean that it does not exist. I am glad that math was not done by me, but how have they proved that some integrals can not be taken? The only thing I know up till now is the Liouville theorem, http://www.sci.ccny.cuny.edu/~ksda/PostedPapers/liouv06.pdf. But this is a generalization. | 265780 | How to determine with certainty that a function has no elementary antiderivative?
Given an expression such as [imath]f(x) = x^x[/imath], is it possible to provide a thorough and rigorous proof that there is no function [imath]F(x)[/imath] (expressible in terms of known algebraic and transcendental functions) such that [imath] \frac{d}{dx}F(x) = f(x)[/imath]? In other words, how can you rigorously prove that [imath]f(x)[/imath] does not have an elementary antiderivative? |
329045 | Prove that [imath] x \in Z(G) \to G \ncong \text{Inn}(G) [/imath].
Let [imath]G[/imath] be a group and [imath]x, a \in G[/imath] where [imath]x=a^2[/imath]. Then [imath]x\in Z(G) ⟹ G≇ Inn(G)[/imath]. I have already shown that if [imath]x=b^{-1}a[/imath], then [imath]G≇Inn(G)[/imath] since [imath]\phi_{a}=\phi_{b}[/imath]. However, I don't think this argument applies for the cause that [imath]x=a^2[/imath]. Does anyone have any hints on proving this using properties of isomorphisms? | 324398 | Prove that [imath]G \cong\mathrm{Inn}(G)[/imath] if and only if [imath]Z(G)[/imath] is trivial
Claim: Let [imath]G[/imath] be a group. Prove that [imath]G \cong\mathrm{Inn}(G)[/imath] if and only if [imath]Z(G)[/imath] is trivial. Could anyone offer a hint on proving this claim just using simple properties of group isomorphisms? (i.e., not using the fact that the quotient group [imath]G / Z(G)[/imath] is isomorphic to the group of inner automorphisms of [imath]G[/imath].) EDIT: It turns out that this Claim is false as stated. There are several counterexamples, several of which are provided in answers below, for the case of G being infinite. However, the claim holds for finite groups. |
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